f
A COURSE OF
PURE GEOMETRY
CAMBRIDGE UNIVERSITY PRESS C. F.
CLAY, Manager
LONDON
EDINBURGH
Fetter Lane, E.C. 4
100 Princes Street
NEW YORK: G. P. PUTNAM'S SONS BOMBAY, CALCUTTA, MADRAS: MACMILLAN AND
CO., Ltd.
TORONTO J. M. DENT AND SONS, Ltd. TOKYO: THE MARUZEN-KABUSHIKI-KAISHA :
All rights reserved
A COURSE OF
PURE GEOMETRY CONTAINING A COMPLETE GEOMETRICAL TREATMENT OF THE PROPERTIES OF
THE CONIC SECTIONS
BY
H.
E.
ASKWITH,
D.D.
Rector of Dickleburgh, Norfolk formerly Chaplain of Trinity College, Cambridge
Cambridge at
the University Press
1917
h
First Edition 1903
Reprinted
1911
Netc Edition
1917
n X.
PREFACE npHIS
work
is
a revision and enlargement of ray Course of
Pure Geometry published former edition in that
in
It differs from the
1903.
does not assume any previous know-
it
ledge of the Conic Sections, which are here treated ah
initio,
on
the basis of the definition of them as the curves of projection of a circle. in
This
is
not the starting point of the subject generally
The curves
works on Geometrical Conic Sections.
usually defined by
means
of their focus
and
and their other properties are evolved therefrom. focus and directrix propert}^
are
directrix property,
Here the
established as one belonging to
is
the projections of a circle and
is
it
freely used,
but the
fact
the conies are derived by projection from a circle and
that
therefore possess all its projective properties in the
mind
Many
is
kept constantl}^
of the student.
of the properties of the Conic Sections which can
only be established with great labour from their focus and directrix property are proved quite simply
derived directly from the
Nor
is
indicated
first.
(p. xii)
But
it is
this
is
true that certain ground has to
not very extensive and I have
the few articles which a Student should master
before he proceeds to Chapter ix.
ledge
are
the method employed here any more difficult than
the prevalent one, though
be covered
when the curves
circle.
of cross
ratios,
harmonic />
Without a certain know-
section,
A /\ C\ *\ *%
involution
and the "
elementary principles of conical projection no one can follow
But these things
the argument here adopted.
are
and the advantage gained by the student who from the beginning sees the Conic Sections whole, as he does when they
more
are presented to his
mind
than compensates
any delay there may be through the short
for
as the projections of a circle,
study of the necessary preliminaries. I
hope
that,
thanks to the efficiency of the Readers of the
Cambridge University
Press, there are not
many
be found in this book.
But
I shall
if I
may be informed
if
any are found
misprints to
be grateful
of the)u at the address given below. E.
DiCKLEBURGH ReCTORY, SCOLE,
Norfolk. September 1917.
H. A.
CONTENTS CHAPTER
I
SOME PROPERTIES OF THE TRIANGLE
CONTENTS
Vlll
CHAPTER V CEOSS-EATIOS PAGE
.....
Definition
Twenty-four cross-ratios reducible to Projective property of cross-ratios
six
Equi-cross ranges and pencils mutually projective
CHAPTER
....
47
48
50 54
VI
PEESPECTIVE
............
Definition
62
Triangles in perspective
64
CHAPTER
VII
HARMONIC SECTION Definition and properties of harmonic ranges
Harmonic property Harmonic property
58
'.60
Ranges and pencils in perspective nomographic ranges and pencils
.... .... ...
and polar of circle of quadrilateral and quadrangle
of pole
CHAPTER
71
74 75
VIII
INVOLUTION Definition
and
80
criterion of involution range
83
Involution projective Involution' properties of the circk'
.
.
.
Orthogonal involution Pair of orthogonal X'ays in every involution pencil
.
.
.
.84
....
85 86
CHAPTER IX THE CONIC SECTIONS Definitions
.
Focus and directrix property
........
90 91
Projective projjerties
92
Circle projected into another circle
93
Focus and directrix as pole and polar
94
CONTENTS
IX
PAGE Parallel chords
95
Focus and directrix property established
96 97
(1)
Parabola
(2)
Ellipse
101
(3)
Hyperbola
102
Diameters and ordinates
106
CHAPTER X PEOPEBTIES COMMON TO ALL CONICS and tangent with
Intersection of chord
directrix
....
108
Curves having focus and directrix property are the projections of a circle
110
Pair of tangents.
111
The Normal
113
Latus rectum Carnot's theorem
114
Newton's theorem
117
Some
116
US
applications
Circle of curvature
Conic through
foiu-
120 points of a quadrangl
121
CHAPTER
XI THE PARABOLA
Elementary properties Tangent and normal
126
.
.
.
.
.
.
.
127
Pair of tangents
130
Parabola escribed to a triangle
132
Diameters
134
Circle of curvature
138
CHAPTER
XII
THE ELLIPSE Sum
of focal distance;
constant
144
Tangent and normal
145
Pair of tangents
150
Director circle
.
150
Conjugate diameters
151
Auxiliary circle
153
.
Equi conjugate diameters
156
Circle of curvature
158
CONTENTS
X
CHAPTER
XIII
THE HYPERBOLA I'ACiE
Form
of curve
163
Difference of focal distances constant
164
Tangent and normal
164
On
the length of the conjugate axis
Pair of tangents
.
.
.
.
.166
.
.
168
.
Director circle
169
The conjugate hyperbola
170
Asymptotic properties
171
Conjugate diameters
174
Circle of curvature
.
.
.
.
.
.
.
.187
.
.
CHAPTER XIV THE EECTANGULAR HYPEEBOLA Conjugate diameters
192
Perpendicular diameters
.
,
.
Rectangular hyperbola circumscribing a triangle
....
Chord and tangent properties
193
194 196
CHAPTER XV OllTHOGONAL PEOJECTION 201
Principles
Fundamental propositions
The
ellipse aa orthogonal projection of a circle
....
202 205
CHAPTER XVI CROSS-EATIO PEOPEETIES OF CONICS
P
constant
210
theorem Brianchon's theorem
214
{ABCD)
Pascal's
Locus of centres of conies through four points Involution range on a conic
....
215
215
.216
CHAPTER XVII EECIPEOCATION 220
Principles
Involution properties of quadrangle and quadrilateral
Desargues' theorem and
its reciprocal
.
.
.
224 227
CONTENTS •
XI
....... .... ....... ..........
Reciprocation applied to conies
Special case where the base conic
is
a circle
PAGE
228 231
Coaxal circles reciprocated into confocal conies
234
A
236
pair of self-conjugate triangles
Reciprocal triangles
237
CfTAPTER XVIII CmCULAE
POINTS.
FOCI OF CONICS
........ ......... ........ ...... .......
Definition of circular points
Analytical point of view
Properties of conies obtained by using circular points
The
Two
four foci of a conic
.
.
.
triangles circumscribing a conic
Generalising by projection
242 243
244 246 248 249
CHAPTER XIX INVERSION
........... ..........
Inversion of line and circle
Inversion of sphere
.
.
256 258
Inversion of inverse points into inver.se points
259
Feuerbach's theorem
262
CHAPTER XX SIMILARITY OF FIGURES
.......... ......... .......'
Horaothetic figures
but not homothetic
Figiu-es directly similar
Circle of similitude for
two
circles
267
270 271
Figures inversely similar
272
Miscellaneous E.xamples
276
Index
285
J
The student who may be using book on Geometrical Conic Sections
Chapter
IX
this will
work as a be able
firsi?' t'
.
xl
to j^ro^^^r" to
after reading the following paragraphs of tW'i'Tst
eight chapters
13 to 16a, 29 to 35, 40 to 45, 48 77 to 87.
to 53, 58, 67, 68,
69
^-
"'^
CHAPTER
I
SOME PROPERTIES OF THE TRIANGLE Definition of terms.
1
iy
^1/(66%
unless otherwise stated, will be
meant
straight
4.
The
,
'
(c)
lines joining the vertices of a triangle to the
the opposite sides are called
its
xhe centre of
middle
medians.
the circumcircle of a triangle
'y
through
•^ps<5ing
its
is
meant the
circle
vertices.
this circle will be called the circiimcentre of
the triangle.
The reader already knows
that the
circumcentre
is
the
point of intersection of the perpendiculars to the sides of the
drawn through
tiiangle
The
(a)
their middle points.
incircle of a triangle is
sides of th? triangle
the circle touching the
and lying within the
triangle.
The centre of this circle is the incentre of the triangle. The incentre is the point of intersection of the lines bisecting the angles of the triangle.
An
(e)
ecv.xle of *"'
..tie
An (•i.e
a triangle
is
a circle touching one side of
the other fewo sides produced.
ecentre
of an ecircle is
is
called
an
There are three
ecentre.
the point of intersection of the bisector of
of tb^ angles and of the bisectors of the other two external
angles. A.
G.
1
Sd.VK, PPOI'ERTIES
1^
(/)
Two
OF THE TRIANGLE
triangles which are such that the sides
of the one are equal respectively to the sides
and angles and angles of the
other will be called congruent. If
ABC be
congruent with A'B'C, we shall express the
by the notation
Proposition.
2.
a triangle on
to
orthocentre)
and
;
fact
A ABO = A A'B'C.
:
'The
peiyendiculars
the opposite sides the distance
from
the vertices
meet in a point (called
of tJie
of each vertex from the ortho-
centre is trvice the perpendicular distance of the circumcentre
from
the side opposite to that vertex.
Through the
vertices of the triangle ^Z?6' draw lines parallel
to the opposite sides.
similar to the triangle
The
triangle
ABC, and
A'B'C
thus formed will be
of double its linear dimensions.
Moreover, A, B, G being the middle points of the sides of A'B'C, the perpendiculars from these points to the sides on which they lie will meet in the circumcentre of A'B'C.
But these perpendiculars are A, B,
C
also the perpendiculars from
to the opposite sides of the triangle
Hence the
first
part of our proposition
is
ABC proved.
SOME PROPERTIES OF THE TRIANGLE
Now
P
let
3
the circumcentre of
be the orthocentre and
ABC. Draw OD perpendicular Then
since
P is
to EC.
PA
and OD are corresponding A'B'C, ABC.
Hence 3.
^P is
A'B'C. two similar triangles
also the circumcentre of the triangle lines in the
twice OD.
Definition.
It
will
be convenient to speak of the
perpendiculars from the vertices on to the opposite sides of a triangle as the peiyendiculars of the triangle; and of the perpendiculars from the circumcentre on to the sides as the
perpendiculars from the circumcentre. 4.
Prop.
The
circle
through the middle points of the sides
of a triangle passes also through the feet of the perpendiculars of the triangle and through the middle points of the three lines joining the ortliocentre
Let D, E,
ABC, centre,
L,
P
M,
N
to the vertices
of the triangle.
the middle points of the sides of the triangle
the feet of
its
perpendiculars,
the circum-
the orthocentre.
Join FD,
Then
F be
DE, FL, LE.
since
E
is
the circumcentre of
ALC,
/.ELA= ZEAL. 1—2
SOME PROPERTIES OF THE TRIANGLE
And
.-.
a like reason
for
ZFLA= ZFAL. zFLE= zFAE = Z FDE since AFDE is a parallelogram. .•.
Similarly
L
is
M and N
lie
on this
Further the centre of this lines bisecting
DL, EM,
FN at
circle.
circle lies
on each of the three
right angles.
Therefore the centre of the circle of
DEF.
on the circumcircle of
at
is
U the
middle
point,
OP.
Now join
DU and
The two
triangles
gr uent, so that
UD =
produce
it
meet
to
AP
in
PUX are easily UX and XP = OD. OUD,
X.
seen to be con-
X lies on the circle through D, E, F, L, M, N. since XP ^ (jD = \AP, X is the middle point of AP.
Hence
And
Similarly the circle goes through of
BP
Thus our proposition 5.
Y and
Z, the middle points
and CP.
The
circle
of the triangle.
is
proved.
thus defined
Its radius
is
is
known
as the nine-points circle
half that of the circumcircle, as
obvious from the fact that the nine-points circle circle of
DEF, which
is
similar to
ABC
is
is
the circum-
and of half
its linear
SOME PROPERTIES OF THE TRIANGLE
Or the same may be seen from our
diuiensions.
DX = OA,
ODXA
for
is
5
figure wherein
a parallelogram.
It will be proved in the chapter
points circle touches the incircle
on Inversion that the nineand the three ecircles of the
triangle.
If the perpendicular AL of a tiiangle meet the circumcircle in H, then PL = LH,
Prop.
6.
produced
to
ABC he P being
the orthocentre.
Join
BH.
Then Z
HBL = z HAC in the same segment = Z LBP since each is the complement
of
/LACB.
Thus the
triangles
PBL, HBL have
also their right angles at
L
their angles at
equal, and the side
BL
B equal,
common.
\-.PL = LH. 7.
Prop.
The feet of
the perpendiculars fi^om
on the circumcircle of a triangle
ABC
any point Q
on to the sides of the
triangle are collinear.
Let R, figure.
S,
T
be the
Join QA, QB.
feet of the perpendiculars as in the
SOME PROPERTIES OF THE TRIANGLE
H
Q'fAS
a cyclic quadi'ilateral since
is
T and S
are right
angles. .-.
^AT>S = zAQS
= complement = complement
QAS zQBG
of Z of
(since
QAiJ,
QBC
aie
supplementary)
= z BQR = z BTR .'.
RTS
This line
known
also as the
The converse If
is
is
QBRT
is cyclic).
a straight line.
called the pedal line of the point Q.
Simson
a triangle are
perpendiculars collinear,
Q
from a
lies
point
viz.
Q
QBRT and QTAS are cyclic, BQR = z BTR = zATS = z AQS. z QBR = z QAS, so that QBCA is cyclic. z
.-.
on
to
the
on the circumcircle of the
triangle.
For since
It is
line.
of this proposition also holds good,
the feet of the
sides of
(since
RTS
SOME PROPERTIES OF THE TRIANGLE Prop. The pedal line of 8. P, the orthocentve of the triangle.
Join
QP
Join
bisects the line joining
Q
cutting the pedal line of
Let the perpendicular Join
Q
AL
meet the
QH cutting the pedal PN and QB.
Tlien since
QBRT
is
zQRT^zQBT = Z QUA = Z HQR .'.
.-.
J/
i>
line in
in
circunicircle in
M and BC in
in
QR
is
parallel to
since
QN.
A 7\VZ = A HNL
is
parallel to
/?7'.
QK'.KP = QM.MN. QK = KP. .-.
J H.
QM = iMR.
= z RNM = z i¥iei\^.
.-.
N.
same segment
since
the middle point of
PA^
K.
cyclic,
But Z PiVZ .= z ZiV^
.-.
7
H.
Q
to
SOME PROPERTIES OF THE TRIANOLK
8 9.
and
Prop.
The
this point is
three luedians of a triangle meet in
a point of
trisection
the line joining the circumcentre
a point,
of each niedian, and also of
and
the orthocentre P.
Let the median
AB o{
Then from the
similarity of the triangles
the triangle
ABC cut OP
in G.
GAP, GDO, we
AP = 20D, that AG = 2GB and PG = 2G0. the median AD cuts OP in G which is a point
deduce, since
Thus
of
trisection of both lines.
Similarly the other medians cut
which
will
OP
in the
be a point of trisection of them
This point
The reader
G
is
called the
same point
G,
also.
median jmint of the
triangle.
probably already familiar with this point as the centroid of the triangle. 10.
is
Prop.
If AD
he a
median of
the triangle
ABC,
then
AB' + AC^^= 2 AD' + 2BD\ Draw Then and
B
AL
perpendicular to BC. AG-' = A B' + BC - 2BC BL AD' = AB' + BD'-2BD.BL. .
These equalities include the cases where both the angles and C are acute, and where one of them, B, is obtuse, provided
SOME PROPERTIES OF THE* TRIANGLE that
BG
BL
and
9
be considered to have the same or opposite
signs according as they are in the
same or opposite
directions.
Multiply the second equation by 2 and subtract from the first,
then
AC' - 2 AD' = . •
.
= The
11.
BC - AB' - -IJWi + BC - 2 BD'
AB'+ AC' = 2
/)-
2AD''
+
2BD-, since
proposition proved in the last article
special case of the following general one //'
lU)
=
D ^
BC = 2BD.
he a point in the side
BC
then
(n
- 1)
BC
is
only a
:
ABC
of a triawjle
such that
n
AB"-
vAC- = n. AD' +
For proceeding as before,
if
(l
-
BC.
^)
we now multiply the second first we get
of
the equations by n and subtract from the
.-.
A C -n.AD'={\-n)AB' + BC -
n
(71 -\)AB' + AC'=n. AD' + BC -
n
.
BD\
(^
BC
= n.AD' + (l-^\BC 12.
incircle
A,
B,C
Prop.
The distances of the points of contact of the ABC ivith the sides from the vertices
of a triangle are
s
— a,
s
— b,s — c
the points of contact
of the
respectively;
and
ecircle opposite to
the distances of
A
are
s,
s
—
c.
SOME PROPERTIES OF THE TRIANGLE
10 .V
to
-b A
respective/ 1/ ,
B,
G and
s
;
a, b, c
being the lengths of the sides apposite
half the sum of them.
Let the points of contact of the incircle be L, M, N.
Then
AM=AN', CL = CM and BL = BN,
since .•.
AM+ BC = h&,U the sura of AM = -a. .-.
Similarly
Next
BL^BN^s-b,
let L',
M',
the sides
=
s,
s
N' be
and
GL = CM=s-c.
the points of contact of the ecircle
opposite to A.
Then and
AN' = AB-\-BN' = AB + BL' AM' = AG + GM' = AG + GL. since AM' =^AN', 2AN' = AB + AG+BG=2s. AN' = s, BL' = BN' = s-c, and GL' = GM' =s-b. •
.-.
.'.
and Cor.
BL'
middle point.
=
GL, and thus LIJ and
BG
have the same
SOME PROPERTIES OF THE TRIANGLE
EXERCISES Defining the pedal triangle as that formed by joining the shew that the pedal triangle
1.
feet of the perpendiculars of a triangle,
has for its
its
incentre the orthocentre of the original triangle, and that
angles are the supplements of twice the angles of the triangle.
A
2.
straight line
PQ
P
the pedal lines of
is
drawn
ABC
i)arallel
AB P and
to
to meet the
shew that and Q intersect on the perpendicular from C
circumcircle of the triangle
in the points
Q,
on AB.
Shew
3.
tlie pedal lines of three points on the circumcircle form a triangle similar to that formed by the three
that
of a triangle
points.
The
4.
j^edal lines of the extremities of
a chord of the circum-
a triangle intersect at a constant angle. the middle point of the chord. circle of
Given the circumcircle
5.
prove that the
loci of its
of a triangle
Find the locus of
and two
of its vertices,
orthocentre, centroid and nine-points centre
are circles.
The
6.
squares of
locus of a point which its
is
such that the sum of is constant
distances from two given points
tlie is
a
sphere. 7.
triangle
A\ B\ C are three points on the sides BC, CM, AB of a ABC. Prove that the circumcentres of the triangles ABC,
BC'A', CA'B' are the angular points of a triangle which to
is
similar
ABC.
A
8.
triangle
AB CA,
circle is described concentric
^^C, and
it
with the circumcircle oi the
intercepts chords A^A.,, B^Bo, C^Co on
BC,
('A,
respectively; from Ai perpendiculars A^b^, A^c^ are drawn to respectively, and from A.^, i^,, B.,, C,, Co similar perpen-
AB
are drawn.
diculars
triangles, of
Shew that
which Ab^Ci
with the nine-points
is
circle,
the
circumcentres of the si.\ lie on a circle concentric
a typical one,
and
of radius one-half that of the original
circle. 9.
A
plane quadrilateral
points
(i)
is
divided into four triangles
shew that the quadrilaterals having the orthocentres and (ii) the circumcentres
internal diagonals
;
b\- its
for angular of the foui-
SOME
12
I'H(>pp:hties
of the triangle and
triangles are similar parallelograms;
A
and
Ao,
Prove that the
10.
if
a triangle to that
farthest from the base passes
is
through the point of contact of the escribed
vei'tex
and
A^ = 4Ao.
line joining the vertex of
point of the inscribed circle which
11.
their areas be Aj
2A +
be that of the quadrilateral, then
circle
with the base.
magnitude and position the lines joining the of a triangle to the points in which the inscribed circle and Given
in
the circle escribed to the base touch the base, construct the triangle.
Prove that when four points A, B,
12.
equal
C,
D
lie
BCD, CDA, BAB,
orthocentres of the triangles
on a
ABC
circle, lie
the
on an
circle.
Prove that the pedal
13.
lines of the extremities of a
diameter
on the
of the cii'cumcircle of a triangle intersect at right angles
nine-points circle.
ABC
14.
to
BC
is
a triangle,
its
circumcentre
meets the circumcircle in K.
perpendicular to
AK
will bisect
KP,
OB
;
Prove that the
P
perpendicular
line
through
B
being the orthocentre.
Having given the circumcircle and one angular point of a and also the lengths of the lines joining this point to the orthocentre and centre of gravity, construct the triangle. 15.
triangle
AB
If
16.
be divided at
and
if
P
manner that
in such a
AO = m.
I.
OB,
be any point, prove /
.
AP-' +
If a, b, c
m
.
BP- =
{l
+ m) OP- +
I
.
AO' + m
.
be the lengths of the sides of a triangle
locus of a point
P
such that a
.
PA^ +
b
.
PB'^ +
c
.
PC"
BO'.
ABC, is
find the
constant.
13
CHAPTER
II
SOME PROPERTIEH OF CIRCLES
.
two points
radius of a circle whose centre
same
When
Definition.
13.
saine
side of
P
and F' lie on the and are on the
is
and their distances from
OP' = square
are
such that
of the radius, they are called inverse points
with respect to the
circle.
The reader can already prove
fur
himself that
P
tangents be drawn from an external point
if
a pair of
to a circle, centre
0, the chord joining the points of contact of these tangents
at right angles to
OP
OP, and cuts
in
a point
which
is
is
the
inverse of P. 14.
The
following proposition will give the definition of the
polar of a point with respect to a circle
:
Prop. The locus of the points of intersection of pairs of tangents drawn at the extremities of chords of a circle, which pass through a fixed point, point,
and
Let
A
is
a straight
line,
the point is called the pole
be a fixed point
in the
called the polar of that
of the
line.
plane of a
circle,
centre 0.
Draw any chord QR
of the circle to pass through A.
Let the tangents at
Q and
R
meet
PL perpendicular to OA. OP cut QR at right angles
in P.
Draw Let
in
M.
SOME PROPERTIES OF CIRCLES
14
PMLA
Then
.-.
.•.
/y is
Thus the and cutting
is cyclic.
OL.OA = OM .OP = sqnAYO a fixed
locus of
it
is
of radius.
the inverse of A.
a straight line perpendicular to
OA,
in the inverse point of A.
It is clear
15.
F
})()int, viz.
from the above that the polar of an external
point coincides with the chord of contact of the tangents from that point.
And
if
we introduce
the notion of imaginary lines,
with Avhich Analytical Geometry has furnished
us,,
we may say
that the polar of a point coincides with the chord of contact of
tangents real or imaginarj^ from that point.
We may is
remark here that the polar of a point on the
circle
the tangent at that point.
Some
writers define the polar of a point as the chord of
contact of the tangents drawn from that point define
by means of
it
in a later chapter.'
harmonic property, which
It
is
The present writer method he has here adopted is the best. Prop.
16.
B BL
Let
Draw
//
others again will
be given
unfortunate that this difference of
treatment prevails.
polar of
;
its
the polar of
A
is
of opinion that the
goes through B, then the
goes through A.
be the polar of
A
AM at right angles
cutting 0x4 at right angles in L. to
OB.
SOME PROPERTIES OF CIRCLES
OM.OB=OL.OA = sq.
Then
.-.
that
A
is.
Two
AM
\s
15
of radiiu
the polar of B,
on the polar o{ B.
lies
points such that the polar of each goes throuoji
thi'
other are called conjugate points.
The reader
will
see for himself that inverse points with
respect to a circle are a special case of conjugate points.
We /,
VI
of
leave
it
as an exercise for the student to prove that if
be two lines such that the pole of
m
will lie
Two
on
/
lies
on
ni,
then the pole
/.
such lines are called conjuc/ate
lines.
From the above property for conjugate points we see that number of coUinear points all pass through a common point, viz. the pole of the line on which they lie. For the polars of a
if
A, B,
C,
D, &c., be points on a line
the polar of
A, B,
We is
C,
P
p whose
goes through A, B, C, kc,
pole .'.
is
P
;
since
the polars of
kc, go through P.
observe that the intersection of the polars of two points
the pole of the line joining them.
SOME PROPERTIES OF CIRCLES
IG
Prop.
16a.
//"
OP
OQ
and
OPQ
is
such that each vertex
the centre of
a pair of conjugate lines of P and Q, then the triangle pole of the opposite side, and
be
a circle whic/t meet the polar of
in
is the
circle is the orthocentre
tJie
of the triamgle.
OQ must lie on the polar of 0, and it also OP and OQ are conjugate lines. Thus OQ is
For the pole of on OP, since
lies
the polar of P.
Similarly
OP
is
the polar of Q.
Q
C
Also the lines joining
the centre to 0, P,
dicular respectively to the polars of those points,
C
is
are perpen-
Q
and therefore
the orthocentre of the triangle. 17.
circle
Prop.
If
whose centre
P
and Q
is 0,
he
any two
poiiits in the
plane of a
then
OP OQ=perp. from P :
on polar of
Q
:
perp.
from Q on
polar of P.
Let P' and Q' be the inverse points of which the polars of P and Q pass.
P
and
Q,
through
PM and QN
Let the perpendiculars on the polars be perp. to OQ and OP respectively.
PT and QR
Then we have since each
is
OP OP' = OQ .
.
OQ',
the square of the radius, and
OR OP = 0T.OQ .
OQ'
since
PRQT is
cyclic,
_OP_OZ- OQ' -or ^~ PM
'OP'~OQ~OR~OP' - OR QN
'
;
draw
SOME PROPERTIES OF CIRCLES Thus the proposition
is
p
proved.
17
SOME PROPERTIES OF CIRCLES
18
Since points on the
common
chord produced of two inter-
secting circles are such that tangents from circles are equal, circles
we
goes through their
common
points
to the two two intersecting
them
see that the radical axis of
And
introducing
the notion of imaginary points, we may say that the radical axis of two circles goes through their common points, real or imaginary. 19.
coplanar
The difference of the squares of the tangents to two circles, from any point P in their plane, varies as the
perpendicular
Let centres
from
P
PQ and PR A and B.
PN
on their radical axis.
be the tangents from
be perp. to radical axis NL, and Let be the middle point of AB. Join PA, PB.
P
to
the circles,
PM to AB;
let
Then
PQ'- PR' = PA' -AQ'- (PB' - BR') = PA' - PB' - AQ' + BR' = AM'-MB'-AQ' + BR'
^203I.AB-20L.AB = 2AB.LA[=2AB.NP.
(see § 18)
SOME PROPERTIES OF CIRCLES
19
This proves the proposition.
"
We may mention here that some writers use the term power of a point " with respect to a circle to mean the square
of the tangent from the point to the circle.
Prop.
20.
three coplanar circles taken
The radical axes of
in pairs meet in a point.
Let the radical axis of the circles
A
and C
circles
A
and
B
meet that of the
in P.
2—2
SOME PROPERTIES OF CIRCLES
20
Then the tangent from
P
= tangent
to circle
from
= tangent from .•
21.
.
P is
Coaxal
P P
G
to circle B.
on the radical axis of
circles.
A
A
to circle
B
and
C.
system of coplanar
that the radical axis for any pair of
them
is
circles
the same
is
such called
coaxal.
Clearly such circles will
same straight
all
have their centres along the
line.
Let the common radical axis of a system of coaxal
circles cut
their line of centres in A.
Then the tangents from
A
to all the circles will
be equal.
Let L, L be two points on the line of centres on opposite sides of A, such that AL, AL' are equal in length to the tangents from A to the circles L and L' are called the limiting ;
points of the system.
They
,
are such that the distance of any point
radical axis from either of
tangent from
P
to the
them
system of
is
P
on the
equal to the length of the
circles.
SOME PROPERTIES OF CIRCLES For radius
if
C
21
be the centre of one of the circles which
is
of
r,
PD = PA' + AD = PA^' +AG''-r''- = PC - r^ = square The two
points
circles of infinitely
They
system.
of tangent, from
P
to circle
C
L
and L' may be regarded as the centres of small radius, which belong to the coaxal
are sometimes called the i^omt circles
of the
system.
The student
will have no difficulty in satisfying himself that two limiting points one is within and the other without
of the
each circle of the system. It
must be observed that the limiting points are
real only in
the case where the system of coaxal circles do not intersect in
For if the circles intersect, A will lie within them and thus the tangents from A will be imaginary. Let it be noticed that if two circles of a coaxal system intersect in points P and Q, then all the circles of the system pass through P and Q. real points. all
Prop.
22.
are inverse points
L
liniitiny points
ivith
of a system of coaxal
respect to every circle of the
circles
si/stein.
C be the centre of one of the circles of the system. and L' be the limiting points of which L' is without the
Let
Let
The
circle C.
SOME PROPERTIES OF CIRCLES
22
Draw tangent L'T
to circle
C; this
will
be bisected by the
radical axis in P.
TN perpendicular
Draw
to line of centres.
L'A:AN=L'F:PT, L'A = AN,
Then
.-.
N coincides with
.'.
Thus the chord
L.
of contact of tangents from L' cuts the line
of centres at right angles in L.
L
Therefore
and L' are inverse
The student
23.
will find it quite easy to establish the
following propositions
Eve7y
two
:
passing through the limiting points cuts
all the
of the system orthogonally.
circles
A
circle
points.
common
tangent
to
two circles of a coaxal system subtends
a right angle at either limiting jmint.
Common
24.
tangents to two
In general four coplanar
Of
common
circles.
these two will cut the line joining their centres ex-
ternally; these are called direct will cut
circles.
tangents can be drawn to two
common
tangents.
the line joining the centres internally
transverse
common
;
And two
these are called
tangents.
P
We
shall
now prove
that the
common
tangents of two circles
cut the line joining their centres in ttuo points which divide that line internally
and externally
in the ratio of the radii.
Let a direct common tangent PQ cut the Join AF, BQ. centres A and B in 0.
line joining the
SOME PROPERTIES OF CIRCLES
Then
BQO
P
since
28
and Q are right angles, the triangles
APO,
are similar,
.-.AO-.BO^AP.BQ. Similarly, if
AB
we can prove AO' O'B
in 0',
We
P'Q' be a transverse common tangent cutting :
= ratio
have thus a simple construction
tangents,
viz. to
AB
divide
circle
;
drawing the common
internally and externally at 0'
and then from
in the ratio of the radii,
tangent to either
of the radii.
for
and 0'
to
and draw a
this will be also a tangent to the other
circle.
If the circles intersect in real points, the tangents from 0'
be imaginary.
will
If one circle
both 25.
and 0'
wholly within the other, the tangents from
lie
will
be imaginary.
Through the point 0, as defined at the end of the last let a line be drawn cutting the circles in RS and
paragraph,
R'S' as
in the figure.
Consider the triangles
We
also the angle at
angles at
R
OAR, OBR'.
OA:OB = AR: BR',
have is
and R'
Thus the
common
is less
to b(jth,
and each of the remaining
than a right angle.
triangles are similar,
and
OR:OR'=AR:BR', the ratio of the radii.
In like manner, by considering the triangles OAS, OBS', in which each of the angles S and *S" is greater than a right angle,
we can prove that OS:
0/Sf'
= ratio
of radii.
SOME PROPERTIES OF CIRCLES
24
We circle
B
thus see that the circle
could be constructed from the
A
by means of the point of all the points on the
from
by taking the radii vectores A and dividing these in
circle
the ratio of the radii.
On
account of this property
called a centre of similitude
is
of the two circles, and the point
R'
is
said to correspond to the
point B.
The student can prove
for
himself in like manner that 0'
is
a centre of similitude.
In order
26.
some given law
is
to
prove that the locus of a point obeying
a circle,
it is
often convenient to
make use
of
the ideas of the last paragraph.
we can prove
If
a
a given
circle, in
be a
circle,
P
that our point
is
such as to divide the
to a varying point Q,
line joining a fixed point
which describes
then we know that the locus of
ratio,
which with the
circle
on which
Q
lies
has
P must for
a
centre of similitude.
For example, suppose we have given the circumcircle of a and two of its vertices, and we require the locus of the
triangle
It is quite easy to prove that the locus of
nine-points centre.
the orthocentre
is
a circle, and from this
locus of the nine-points centre
circumcentre (which
is
describes a circle) and
and
OU = \0P
;
a
given) and
U the
it
follows that the
be the
circle, since, if
P
the orthocentre (which
nine-points centre,
therefore the locus of
centre in the line joining
P
is
U is
U lies
on
a circle, having
to the centre of the circle
OP its
on which
lies.
27.
Prop.
that its distances
constant ratio
A
is
The locus of a point which moves in a plane so fixed points in that plane are in a a circle.
from two
B be the two given points. and externally at C and D in the given are two points on the locus. Let
Let
and
P be
any other point on the
Divide ratio, so
locus.
AB internally that C and D
SOME PROPERTIES OF CIRCLES
Then
25
since
AP:P£ = AC:GB=AD:BD, PD are the internal and external bisectors of the
.•.
PC and
.'.
CPD
.•.
the locus of
/-APB.
Cor. is
a circle on
is
CD
D in
P
AB be
subtends a right angle, then
described,
locus
divided internally and externally
PC
P be any point PD are the
and
00' joining the two centres
A and
B, as defined in
follows from § 27 that if
it
its
at
which
internal
CD and
Z APB.
If on the line
28.
of circles, centres
as diameter.
be not confined to a plane,
the same ratio, and
external bisectors of
CB
as diameter.
If the line
2.
C and
P
If the point
1.
the sphere on
Cor. at
a right angle.
is
C
§ 25,
of similitude
a circle be
be any point on this
circle,
CA CB = radius :
The tude.
circle
of
A
circle
on 00' as diameter
is
:
radius of
B
circle.
called the circle of sindli-
Its use will be explained in the last chapter,
treat of the similarity of figures.
when we
SOME PROPERTIES OF CIRCLES
26
EXERCISES P
be any point on a given circle A, the square of the tangent from P to another given circle B varies as the perpendicular If
1.
P from
distance of
any point
C
to
A and B
If ^,
other
circle,
jB,
C
P
a point
are in a given ratio, the locus of 4.
A and
B.
the tangents
circles,
drawn from
are in a constant ratio.
drawn from
If tangents
3.
B
of
the radical axis of
be three coaxal
If A, B,
2.
P
to
two given
circles
&c. be a system of coaxal circles
X
then the radical axes oi A,
;
B,
X
;
C,
and
A and
A and
a circle coaxal with
is
B.
X be any
X &c.
meet in
a point. 5. The square of the line joining one of the limiting points of a coaxal system of circles to a point P on any one of the circles
P
varies as the distance of
two
If
6.
from the radical axis.
two others orthogonally, the radical axis joining the centres of the other pair, and
circles cut
of either pair is the line
passes through their limiting points.
from any point on the
If
7.
given
circles, pairs of
between one pair
is
The three
8.
circle of similitude (§ 28) of
tangents be drawn to both
circles,
two
the angle
equal to the angle between the other pair.
cii'cles
of similitude of three given circles taken
in pairs are coaxal. 9. Find a pair of points on a given two given pairs of points.
If
10.
any
line
cut two given
circle concyclic
with each of
in P, Q and F', Q' which the tangents at F on a circle coaxal with the
circles
respectively, prove that the four points in
and Q cut the tangents given
at
F' and Q'
lie
circles.
11.
A
line
FQ
is
drawn touching
at
P
a circle of a coaxal
system of which the limiting points are K, K', and Q is a point on the line on the opposite side of the radical axis to P. Shew that if T, T' be the lengths of the tangents drawn from F to the two concentric circles of
respectively
which the common centre
QK, QK\ then
T:T' =PK.FK'.
is
Q^
and whose
radii are
SOME PROPERTIES OF CIRCLES
27
a fixed point on the circumference of a circle C, P any other point on C the inverse point ^ of f is taken with respect to 12.
is
;
a fixed circle whose centre
is
at 0, prove that the locus of
Q
is
a
straight line.
Three
13.
circles Cj, Co, C^ are
and
section of Ca
such that the chord of inter-
passes through the centre of C^
and the chord and Cj through the centre of Co shew that the intersection of Cj and C.j passes through the centre of CV C'g
of intersection of C\
chord of
,
;
14. Three circles A, B, C are touched externally by a circle whose centre is P and internally by a circle whose centre is Q. Shew that PQ passes through the point of concurrence of the radical
axes oi A, B,
AB
15.
C
is
taken in
a.
diameter of a
produced,
C
points of
A and B with
any point on
circle S,
a circle whose centre
C
respect to
pairs.
is
respect to C.
AB
or
AB
A' and B' are the inverse
at 0.
Prove that the pole with
of the polar with respect to
S
of the point
is
the
middle point of A'B'.
A
16.
system of spheres touch a plane at the same point 0,
prove that any plane, not through 0, will cut them in a system of coaxal circles.
A
17.
point and
its
polar with respect to a variable circle being
given, prove that the polar of
any other point A passes through a
fixed point B. 18. circles
system
.4 ;
is
a given point in the plane of a system of coaxal
prove that the polars of all
A
with respect to the circles of the
pass through a fixed point.
28
CHAPTER THE USE OF
III
CONCURRENCE AND
SIGNS.
COLLINEARITY The reader
29.
is
already familiar with the convention of
and Analytical Geometry
signs adopted in Trigonometry
measurement of straight lengths measured along a
lines.
in the
According to this convention
from a point are counted positive
line
or negative according as they proceed in the one or the other direction.
With
this
convention we see that,
if
A, B,
C
he three points
in a line, then, in whatever order the points occur in the line,
AB+BC = AG.
AC If
C
lie
in this case
between
in passing from
A
BC is
and B,
AB + BC does
final distance
From
A
B
to B,
of opposite sign to
AB, and
not give the actual distance travelled
and then from
B
to G,
but gives the
reached from A.
the above equation
we get
BC = AC-AB. This reduce
is
all
an important identity.
By means
of
it
we can
our lengths to depend on lengths measured from a
fixed point in the line.
This process
speak of as inserting an origin.
Thus,
it
if
AB^OB-OA.
will
we
be convenient to
insert the origin 0,
THE USE OF
M he the middle point of the
Prop. If any other point
30.
be
CONCURRENCE AND COLLINEARITY
SIGNS.
29
AB, and
line
in the line, then
20M=0A + 0B. O
M
A
B
AM=MB,
For since
we have
by inserting the origin
OM-OA = OB-OM, .-.
A number
31.
20M=0A+0B.
of collinear points are said to form a range.
G, D he a range of four points; AB.CD + BC.AD+CA.BD = 0.
Prop.
If A, B,
BCD
A
For, inserting the origin A,
we
see that the above
AB{AD - AC) + {AG - AB) AD - AC {AD - AB),
==
and
then
this is zero.
This
is
32.
If
an important identity, which we
A, B,
side their line,
C
be a range of points, and
we know that the area
the area of the triangle
bases
shall use later on.
OBC
any point out-
of the triangle
OAB is
to
in the ratio of the lengths of the
AB, BG.
O
Now
BG
if
we
and the
ratio
AB, we cannot substitute for this A 0^6* unless we have some convention respecting are taking account of the signs of our lengths
ratio
A OAB
:
AB BG occurs, :
THE USE OF SIGNS
30
the signs of our areas, whereby the proper sign of
be retained when the
AB BC will :
ratio of the areas is substituted for
it.
is that the area of a triangle PQR be accounted positive or negative according as the triangle
The obvious convention shall is
to the one or the other side as the contour
Thus contour
PQR, we
nitude,
while
shall
APRQ
describing the contour
With
is
C
is
described.
to our left
will
PRQ
this convention
points A, B,
PQR
hand as we describe the consider A PQR to be a positive mag-
the triangle
if
be a negative magnitude, the area
we
is
for
in
on our right hand.
see that in whatever order the
occur in the line on which they
lie,
AB:BC=AOAB:AOBG, = AAOB-.ABOG.
or
It is further clear that with our convention
we may say
AOAB+ AOBC= AOAC, AOAB- AOAC= AOCB,
and
remembering always that A, B, G are coUinear. 33.
triangle
Again, we
OAB
is
know
that the magnitude of the area of a
^OA .OB sin AOB,
and
it is
sometimes con-
make use of this value. But if we are comparing OAB, OBC by means of a ratio we cannot substitute
venient to the areas
^OA .OB sin AOB
I
for
and ^ OB.
OC sin BOG
BPA
them unless we have a further convention
of signs
whereby
the sign and not merely the magnitude of our ratio will be retained.
The obvious convention here again positive if described in one sense
will
be to consider angles
and negative in the opposite
sense; this being effective for our purpose, since sin(— ^r)
= — sin x.
CONCURRENCE AND COLLINEARITY
APB = - Z BPA.
In this case Z
PA, and
P
round
and
BPA
the angle
PB;
from the position
APB is to be P from the revolution of PA
The angle
regarded as obtained by the revolution of position
31
PB
as the
round
these are in opposite senses'
so of opposite signs.
With
this convention as to the signs of our angles
argue from the figures of
we may
§ 32,
AB _ AAOB _ I OA.OBsinZAOB
BC ~ "KbOG ~ ^OB.OCsm^ BOG OC being all regarded as positive) _0A sin z^ OS ~" 00 sin Z BOO
(the lines (JA, OB,
'
•
AB is retained in the process
In this way the sign of the ratio j^p of transformation, since sin
ZAOB
and
sin
Z
50C
same or opposite sign according of the same or opposite sign. are of the
as
AB
and
BC
arr
The student useless
will see that our convention would have been had the area depended directly on the cosine of the
angle instead of on the sine, since cos
{— A)
=+
cos(.I).
Test for coUinearity of three points on the sides of a triangle. 34.
The
following proposition,
known
as Menelaus' theorem,
is
D, E,
F
A, B,
C
of great importance.
The necessary and
sufficient condition that the points
on the sides of a triangle
ABC
opposite to the vertices
respectively should be collinear is
AF.BD.CE = AE.CD.BF, regard being had
to the
signs of these lines.
All these lines are along the sides of the triangle. shall
consider any one of
according as the triangle
we
travel along
it.
is
them
to
We
be positive or negative
to our left or right respectively as
THE USE OF SIGNS
32
We will first prove F are collinear.
that the above condition
is
necessary, if
D, E,
Let
p, q, r
DEF, and
line
be the perpendiculars from A, B, C on to the let these be accounted positive or negative
according as they are on the one or the other side of the line
DEF. With
this convention
we have
CONCURRENCE AND COLLINEARITY
33
..(AF+ FF') BF= AF (BF + FF'), ..FF'(BF-AF) = 0, ..FF' = 0, .'.
F coincides
Thus our proposition
35.
vertices
Test
for
is
with F'.
completely proved.
concurrency
of
lines
through
the
of a triangle.
The following
proposition,
known
as
Ceva's
theorem,
is
fundamental.
The necessary and
suffi,cient
condition
t/uit
the lines
AD, BE,
34
CF
TEIE
USE OF SIGNS
ABC F should be concurrent is AF.BD.CE = -AE.CD. BF,
drawn through
the vertices of a triangle
to
meet the
opposite sides in D, E,
the
same convention of signs being
as in the last
adojyted
proposition.
First let the lines
AD, BE, OF meet
in P.
Then, regard being had to the signs of the areas,
AF_ is AFC _ AAFP _ BF~ ABFC ~ A BFP ~ BD ^ ABDA _ ABDP ^ GD~ A CDA ~ A CDP " CF
_
AE '
AAFC- AAFP _ AAPC ABFC - ABFP ~ ABPG ABDA- ABDP _ ABPA ACDA -A GDP ~ XCPA AGEB _ A CEP _ ACEB - A CEP _ AGPB AAEB ~ AAEP ~ AAEB-AAEP ~ AAPB AF.BD.CE AAPC ABPA AGPB ' AE. CD. BF~ AGFA' AAPB ABPG
'
'
'
= (-l)(-l)(- 1)^-1. Next
D, E,
let
F be
points on the sides of a triangle
such that
AF.BD.CE = -AE. CD then will
AD, BE, CF
B Let AD,
.
BF,
be concurrent.
C
D
BE
meet
in Q,
and
let
GQ meet
AB in
F'.
ABC
CONCURRENCE AND COLLINEARITY ..
35
AF' .BD .CE=-AE.GD.BF'.
Ar_AF •'•
BF'~ BF' FF')BF = {BF + FF')AF. ..FF'(BF-AF) = 0. ..FF' = 0.
.•.(AF +
.'.
F' and
Hence our proposition Prop.
36.
triangle
// D, E,
ABC opposite to AF.BD.GE A E CD BF .
.
is
F coincide.
completely proved.
F
he three points on
A, B,
C
s'mA CF sin sin
the sides of a
respectively.
BA L) sin CBE
A BE sin CA D sin BCF
"
For
with
BD ABAD _^AB.AD s'm BAD _AB sm BAD CD ~ 'KOAD " fAC\~ADliin^UAlJ ~~ AG 'sin GAD' our convention as to sign, and AB, AC being counted
positive.
Similarly
and
'
'
AF AG
^xnACF
BF GE
sin
BlJ'smBZ'F
BG
CBE
AE AB sin ABE' AF.BD.GE _ sin AGF sin BA D sin CBE AE.GD .BF~ sin ABE sin GAD sni BGF
THE USE OF SIGNS
'.]()
The necessary and
Cor.
OF
should be concuiTent
sufficient condition that
AD, BE,
is
BAD sin CBE ~_ _
s in
^Ci^ sin
sin
A BE sin GA D sin BCF A
If
be the point of conciirrence this relation can be written
in the form sin
^^0 sin ^(70 sin CM
sin
Z(X)sin
GBO sin BA6~~
'
being easy to remember.
this
37.
Isogonal conjugates. Two lines AD, AD' through A of a triangle which are such that
the vertex
zBAD = zD'AC (not
z GAD')
are called isogo7ial conjugates.
Prop. the
vertices
// AD, BE, of a
GF
triangle
he three concun-ent lines
ABG,
their
isogonal
AD', BE', GF' will also he concurrent.
For
sin
sin
BAD _ sin D' AG _ sin GAD' GAD ~ sin DAB ~ siu^BAD' sin G BE _ si n ABE '
Similarly
ABE ~ sin GBE' sin AGF _ s in BGF'
sin
and
sin
BGF ~
sin
AGF'
tli
rough
conjugates
CONCURRENCE AND COLLINEARITY
sin
sin
sin ABE' sin BCF BAD' sin CBE' sin ACF'
CAD'
'
sin
5^i) sin
(75£; sin
sin G'^Z> sin ^i?A' sin •.
The
38.
are called
its
37
.4/)', i?A"',
ACF = -1, BGF
CF' are concurrent.
isogonal conjugates of the medians of a triangle
symmedians.
Since the medians are concurrent,
symmedians are concurrent also. The point where the symmedians intersect is called the symmedian point of the the
triangle.
The student
will see that the
concurrence of the medians
and perpendiculars of a triangle follows at once by the tests of this chapter (§§ 35 and 36). It was thought better to prove them by independent methods in the first chapter in order to bring out other properties of the orthocentre and the median point.
We
39.
will
conclude
this
chapter by introducing the
student to certain lines in the plane of a triangle which are called
by some writers antiparallel to the
Let
ABC
be a triangle,
D
and
F
sides.
points
in the sides
THE USE OF SIGNS
38
AB and AC such that zADE = zBCA and therefore also AED = Z CBA. The line BE is said to be antiparallel to BC.
Z
It will
It
may
be
BG are
line
lines antiparallel to
parallel to
an exercise
left as
sym median
the
DBGE
be seen at once that
lines antiparallel to
through
A
and that
is cyclic,
all
one another.
to the student to prove that
of the triangle
ABC bisects all
BG.
EXERCISES The
1
lines joining the vertices of
a triangle to
its
circumcentre
are isogonal conjugates with the perpendiculars of the triangle. 2.
The
lines joining the vertices of a triangle to the points of
contact with the opposite sides of the incircle and ecircles are respectively concurrent. 3.
ABC
is
opposite sides.
If
4.
The midpoints
D and E
:
trisecting
the perpendiculars on the 6' /i
of the sides
U and K. M, and CM point of BE.
AH in
GF
BG
and GA
the trisecting points nearest
respectively are sects
AD, BE,
BH and CK be drawn perpendicular to EF, then AG, BH and will be concurrent.
AG,
DE respectively,
FD,
are
a triangle;
CK
intersects
intersects
BE
B
ABC BC and BA and BL inter-
of the triangle
of the sides
AD
in N.
in L,
Prove that iV
is
a
CONCURRENCE AND COLLINEARITY
39
drawn from the orthocentre of a on the bisectors of the angle A, shew that their feet are collinear with the middle point of BC. If perpendiculars are
0.
ABC
triangle
The points
6.
AB
with suffixes
1, 2,
A, B, or C.
BE.,,
BC
of contact of the ecircles with the sides
of a triangle are respectively denoted
Sit
by the
BC, CA,
letters
D, E,
F
3 according as they belong to the ecircle opposite
CF^
P BE^, CF^ at Q E^F^ and D,E, and AB at Z. Prove that Q and X, Y, Z are respectively
intersect at
X; F^D, and CA
Y
at
;
the groups of points A, P, D^,
;
:
:
collinear.
A and ^are cut in
Parallel tangents to a circle at
7.
the points
C and D respectively by a tangent to the circle at E. Prove that AD, BC and the line joining the middle points oi AE and BE are concurrent.
the angular points of any triangle ABC lines AD, BE, drawn cutting the opposite sides in D, E, F, and making
From
8.
CF
are
equal angles with the opposite sides measured round the triangle in
The
the same direction. Prove that
B'C CA AE BF CD
A' B
.
.
9.
.
.
lines
AD, BE,
CF iorm
CB AF BD.CE
^ A'C^ B'A
.
.
Through the symmedian point
^
a triangle A'B'C.
BC.CA.AB AD. BE. CF
of a triangle lines are
antiparallel to each of the sides, putting the other
two
sides.
that the six points so obtained are equidistant from the
drawn Prove
symmedian
point. circle through these six points has been called the cosine from the property, which the student can verify, that the intercepts it makes on the sides are pi'oportional to the cosines of
[The
circle,
the opposite angles.] 10.
Through the symmedian point
of a triangle lines are
drawn
Prove that the six points so obtaiued are equidistant from the middle point of parallel to each of the sides,
cutting the other sides.
the line joining the symmedian point to the circumcentre.
[The
circle
through these six points
See Lachlan's Modern Pure Geometry,
is
called the
Lemoine
circle.
§ 131.]
AD, BE, CF are three concurrent lines through the vertices ABC, meeting the opposite sides in D, E, F. The circle circumscribing DEF intersects the sides of ABC again in D', E', F\ Prove that AD', BE', CF' are concurrent. 1 1.
of a triangle
THE USE OF
40
12.
CONCURRENCE AND COLLIN EARITY
SIGNS.
Prove that the tangents to the circumcircle at the vertices meet the opposite sides in three points which are
of a triangle
colHnear. 13. If AD, BE, CF through the vertices of a triangle ABC meeting the opposite sides in D, E, F are concurrent, and points D', E', F' be taken in the sides opposite to A, B, C so that DD' and BC, EE' and CA, FF' and AB have respectively the same middle point, then AD' BE', CF' are concurrent. ,
from the symmedian point aS' of a triangle ABC, perpendiculars SD, SE, SF be drawn to tlie sides of the triangle, then S will be the median point of the triangle DEF. 14.
If
15.
Prove that the triangles formed by joining the symmedian
point to the vertices of a triangle are in the duplicate ratio of the sides of the triangle. 16.
The
sides
BC, CA,
internally by points A', B',
AB
BA' :A'C=CB' Also
B'C
produced cuts
BA"
BC :
of
a
triangle
ABC
are divided
C so that :
B'A = AC'
:
externally in A".
CA" = CA"
:
A'B\
C'B.
Prove that
41
CHAPTER
IV
PROJECTKDX 40.
V
If
if
A'
called the projection of
is
A any other meet a given plane tt the plane tt by means
be any point in space, and
VA, produced
then
if
necessary,
A
<»n
point, in ^',
of the
vertex V. It is clear at once that the projection of a straight line
on
TT
namely the intersection of the plane with the plane containing V and the line.
TT
plane, then that line will be projected to infinity
a plane
tt is
a straight
If the plane
plane.
The
line,
through
V and
line thus obtained
a certain lim- be parallel to the
on the
tt
plane
is
o;i
the
tt
called the line
at infinity in that plane.
Suppose now we are projecting points 41. means of a vertex V on to another plane tt. Let a plane through
p
in the line
F parallel
in a plane
to the plane vr cut the plane
AB.
AB will project to infinity on the plane and AB is called the vanishing line on the plane p.
This line this reason
p by
The vanishing
tt,
for
line is clearly parallel to the line of inter-
section of the planes
p and
tt,
which
is
called
tlie
axis
of
projection.
Now let EDF be an angle in the plane p and let its DE and BF cut the vanishing line AB in E and F, then plane into an angle of angle EDF will project on to the
42. lines
the
magnitude EVF.
tt
PROJECTION
42
For
let
Then
intersections of
that
is,
VDE intersect the plane ir in the line de. VEF is parallel to the plane the these planes with the plane VDE are parallel
the plane
since the plane
de
is
parallel to
Similarly df
VE.
parallel to
VF.
Aedf^/.EVF.
Therefore
Hence we
is
tt,
see that
mu/ angle in
the
plane
p projects
on
to the
plane into an angle of magnitude equal to that subtended at V hy the portion of the vanishing line intercepted by the lines
TT
containing the angle. 43. tion,
By
Prop.
any given
line on
a proper choice of the vertex V of projeca plane p can be projected to infinity, while
two given angles in the plane
magnitude on
to
a plane
Let ^J5 be the given Let the plane Let
EDF,
ir
ir
p
are projected into angles of given
properly chosen.
line.
Through
draw any plane
E'D'F' be the angles in the plane
be projected into angles of magnitude
Let E, F,
AB
be taken parallel to the plane
E',
F' he on
AB.
a.
and
/3
p'.
p'.
p which
are to
respectively.
PROJECTION
On EF,
E'F' in the plane
p
containing angles equal to a and
43
describe segments of circles /3
Let these
respectively.
segments intersect in V.
Then
if
V
a
and
AB
be taken as the vertex of projection,
project to infinity,
will
and EDF, E'D'F' into angles of magnitude
respectively (§42).
yS
Cor.
Any
1.
triangle can he projected into
an equilateral
triangle.
For
if
we
project two of its angles into angles of
third angle will project into 60° also, since the
three angles of the triangle in projection
is
sum
60'^
the
of the
equal to tw^o right
angles.
Cor. Let
A quadrilateral can be projected into A BCD be the quadrilateral. Let EF 2.
diagonal, that
is
a square. be
its
third
the line joining the intersections of opposite
pairs of sides.
Let
AC and BD
Now
if
we
intersect in Q.
project
EF
to infinity
and at the same time
44
PROJECTION
project will
Z
s
BA D
and
BQA
into right angles, the quadrilateral
be projected into a square.
For the projection of
EF to
jection shall be a parallelogram
infinity, secures that
;
the projection of Z
a right angle makes this parallek)gram rectangular projection of
Z
AQB
into a right angle
the pro-
BAD into
and the makes the rectangle a ;
square.
It
44.
may happen
preceding paragraph projected to infinity.
EFV is
angle
V will
Suppose that
we must draw
this case
that one of the lines parallel to the line
is
a line
DE is
FV in
the supplement of
be the intersection' of the
a.
line
DE, D'E'
AB
which
parallel to
is
in the
to be
AB.
In
the plane p' so that the
The vertex
FV with
of projection
the segment of
the circle on E'F'. If
D'E'
is
AB, then the vertex F will be the now obtained and another line angle E'F'V is the supplement of yS.
also parallel to
intersection of the line i^Fjust
i^'F so drawn that the 45.
Again the segments of circles described on EF, E'F' in In § 43 may not intersect in any real point.
the proposition of this case
F is
an imaginary point, that
is
to say
it is
a point
algebraically significant, but not capable of being presented to
The notion of imaginary points and lines which we take over from Analytical Geometry into our present subject will be of considerable use. the eye in the figure.
PROJECTION 46.
A
Prop.
45
range of three points
is projective with
any
other range of three points in space.
AV—
C
Let A, B,
/B
be three collinear points, and A', B', C" three
others not necessarily in the same plane with the
Join
Take any point Join VB,
A'
VC
Join DB',
EC.
taining the lines
Let DB',
let
them meet a
;
A'DE diawn
These are in one plane, and A'E.
EC meet in
E
line
ihroucrh o
and E. viz.
the plane con-
A'C
Then by means into A', D,
V in A A'.
and
VAC in D
in the plane
V.
Join V'A'.
of the vertex V, A, B,
C
can be projected
and these by means of the vertex
projected into A', B'
,
V
can be
C.
Thus our proposition 47.
first three.
A A'.
is
proved.
The student must understand that when we speak
of
one range being projective with another, we do not mean necessarily that the one can be projected into the other by a single projection, but that
we can
other by successive projections.
pass from one range to the
PROJECTION
46
A
range o^ four points
is
not in general projective with any
We
other range of four points in space.
shall in the next
chapter set forth the condition that must be satisfied to render the one projective with the other.
EXERCISES Prove that a system of parallel
1.
on
lines in a plane
jy
will project
another plane into a system of lines through the same point.
to
Two
2.
angles such that the lines containing them meet the
vanishing line in the same points are projected into angles which are equal to one another.
Shew that
3.
in general
angles of the same magnitude
Shew that a
4.
plane
a.
triangle can be so projected that
any
line in its
projected to infinity while three given concurrent lines
is
through
three angles can be projected into
its vertices
become the perpendiculars
of the triangle in
the projection. Explain, illustrating by a figure,
5.
lying on a line
PQ, and
jected into a point r lying between of
P
and
6.
how
outside the portion jj
and
q,
it
PQ
is
of
that a point it,
E
can be pro-
which are the projections
Q.
Any
three points ^i, B^, C\ are taken respectively in the
BG, CA, AB of the triangle ABC B^C^ and BC intersect in F; Ci^i and CA in G and A,B, and AB in H. Also FH and BB^ Prove that MG, intersect in A/, and FG and CCj in N. and sides
;
;
NH
BC
are concurrent. 7.
Prove that a triangle can be so projected that three given its vertices become the medians of the
concurrent lines through
triangle in the projection. 8.
If
A Ay,
BBy, CC-^ be three concurrent lines drawn through ABC to meet the opposite sides in A-^B^C^
the vertices of a triangle
and
if
AB m
B-^C-y
Co,;
meet
BC
in A^, C\Ay meet
CA
in
B.^,
and A^B^ meet
then A^, B^, C\ will be collinear.
[Project the concurrent lines into medians.] 9.
If
a triangle be projected from one plane on to another the
three points of intersection of corresponding sides are collinear.
47
CHAPTER V CROSS-RATIOS Definition.
48.
ratio
is
^^„
.
D
1( A, B, C,
he a range of points, the
called a cross-ratio of the four points,
and
is
conveniently represented by (A BCD), in which the order of the letters
the same as their order in the numerator of the
is
cross-ratio.
Some
writers call cross-ratios
'
anharmonic
hoAvever not a fortunate term to use, and
For the term anharmonic '
it.
anharmonic cross-ratio
ratio should be
may
what
ratio of
called
The
49. :
is
an
so that
may be the crossThe student will
to say
when he comes
to
Chapter VII.
essentials of a cross-ratio of a range of four points
(1) that each letter occurs once in both
denominator
is
not harmonic, whereas a
a harmonic range.
better appreciate this point
are
is
This
ratios.'
be best to avoid
will
means not harmonic,
'
one that
be harmonic, that
is
it
(2) that the
;
numerator and
elements of the denominator are
obtained by associating the together,
first and last letters of the numerator and the third and second, and in this particular order.
AB.CD -T-p.
—TTp
—jYn,
^^ stands,
.
•
IS
,
.
,
not a cross-ratio but the negative oi one, tor
though not appearing
be a cross-ratio as
to
becomes one on rearrangement,
,.
.
tor it
=
BA.— CD jyj- y^^ BD CA .
is
(BAGD).
,
,
it
that
CROSS-RATIOS
48
Since there are twenty-four permutations of four letters
taken
all
together,
we
see that there are twenty-four cross-ratios
which can be formed with a range of four points.
Prop.
50.
The twenty -four cross-ratios of a range of four to six, all of tvhich can he expressed in
points are equivalent
terms of any one of them.
{ABGD) =
Let
First
changed
we observe
that
if
\.
the letters of a cross-ratio be inter-
in pairs simultaneously, the cross-ratio is
unchanged.
= (ABGD),
For
= (ABCD), = (ABCD). Hence we get (A BGD) = {BADG) = (GDAB) = {DGBA ) = X...(1). Secondly we observe that a cross-ratio interchange either the
first
and third
is
letters, or
inverted
if
we
the second and
fourth. .-.
{ADGB)==iBGDA)^{GBAD) = {DABG) = l
...(2).
A,
These we have obtained from (1) by interchange of second and fourth letters the same result is obtained by interchanging the first and third. ;
.
A,
49
CROSS-EATIOS
Thus the interchange of the second and third letters changes -\. We may remark that the same result is obtained
into 1
by interchanging the
Thus from
and
first
fourth.
(1)
(ACBD) = (BDAC) = {(JADB) = (DBCA) = and from
this again
-\
1
...
by interchange of second and fourth
{ADBC) = {BCAD) = {CBDA) = (BACB) =
^
In these we interchange the second and third
^
.
(3),
letters,
.
.
(4).
letters,
and
get
(A
BBC) = (BACD) - (GDBA = (IJCAB) )
^
1
And now
interchanging the second and fourth we get
(ACDB) = (IWCA) = (CABD) = (DBAC) = ^-"
^
...
(6).
A,
We
have thus expressed
And we
all
the cross-ratios in terms of X.
see that if one cross-ratio of four coUinear points be
equal to one cross-ratio of four other collinear points, then each of the cross- ratios of the first range
is
equal to the corre-
sponding cross-ratio of the second.
Two
such ranges
51.
Prop.
and D,
may
If A,
E oilier points
be called equi-cross.
C
B,
be three separate collinear points,
in their line such that
(ABCD) = {ABCE), then
D
must coincide
^ Forsmce .
.-.
.-.
E.
luith
AB.CD AB.GE AD7CB = AE~CB' AE.CD = AD.CE.
.-.
AD {CD DE{AD-CD) = 0. DE.AG = 0.
.-.
DE =
{AD + DE) CD = .-.
that
is,
D
and
E
coincide.
for
AC^O,
-f-
DE).
50
CROSS-RATIOS 52.
Prop. A range of four points any plane.
is eqni-cross
with
its
projection on
Let the range
V'mto A' B'G'D'.
A BCD
be projected by means of the vertex
51
CROSS-RATIOS This In sin
is
fig.
obvious in
fig. 1.
2
A'VB' =
sin
B'VA,
= - sin ^ VB, sin A' VD' =
and
these angles being supplementary.
sin
D'VA =
Fig. 3.
Fig. 2.
Further
sin
and
sin
In
fig.
3 sin sin sin
^'F^' = sin.I Fi^, CVD' = sin CVD, A'VD' = sin Z>'F^ = -
sin C'Fi?'
Thus
C"FZ)' = sin CVD, C'VB' = sin CVB.
=
sin
'
BVC =-
sin .4 sin
VD,
CVB.
in each case
{A'B'C'D')
= {ABCD). 4—2
52
CROSS-RATIOS 53.
A number
of lines in a plane which
meet
in a point
V
are said to form a pencil, and each constituent line of the pencil
V is
is
called a ray.
is
called a transversal of the pencil.
Any straight
called the vertex of the pencil.
line in the plane cutting the rays of the pencil
From the last article we see that if VP^, VP.„ VP,, VP^ form a pencil and any transversal cut the rays of the pencil in A, B, C, D, then (ABCB) is constant for that particular pencil; that
is
to say it is
It will
independent of the particular transversal.
be convenient to express this constant cross-ratio by
the notation
V {P,P^P,P,). .0
.,.^v'
.-""Pi
,^''^ P?^'''
Pi"
We
easily see that a cross-ratio of the projection of a pencil
on to another plane
is
equal to the cross-ratio of the original
pencil.
For
let
projection.
V(P^, P., Pj, P,) be the
pencil,
the vertex of
53
CROSS-RATIOS
Let the line of intersection of the p and tt planes cut the be the projection of
rays of the pencil in A, B, C, D, and let
ABCD
Then
V
VP^, and so on.
V, V'P;, of
is
a transversal also of
v'{p;,p:,p:,p:). .-.
54.
V(P,P,P,P,)
We
are
now in
= (ABCD)=
V'{P/P.:P,'P:).
a position to set forth the condition that
two ranges of four points should be mutually projective.
Prop. such
that
//
ABCD
{A'B'C'I)')
be
a range, and A'B'G'D' another range
= (ABCD),
then
the
tivo
ranges
are
projective.
V
B/
V upon VD and let these lines meet VAD in P, Q, R respectively.
Join ^^4' and take any point
it.
Join VB, VC,
A'
in the plane
Join PB', QC and let these meet in V. V'R, the latter cutting A'D' in X.
a line through
Join V'A', and
54
CROSS-RATIOS
Then (ABCD) = (A'PQR) = (A'B'C'X). But
(A BCD)
=
{A'B'C'D') by hypothesis.
..{A'B'G'X)={A'B'G'D'). .'.
X
coincides with
Thus, by means of the vertex V,
A'PQR, and
Def.
Two
is
proved.
ranges
said to be homographic
the one
is
(§ 51).
V into A'B'C'D'.
these again by the vertex
Thus our proposition
55.
D'
ABCD can be projected into
ABCDE... and A' B'C'D'E' ...
when a
cross-ratio of
are
any four points of
equal to the corresponding cross-ratio of the four
corresponding
points
of the
This
other.
is
conveniently
expressed by the notation
(ABCDE...) = {A'B'C'D'E'...).
I
The student will have no difficulty in proving by means of 54 that two homographic ranges are mutually projective.
Two
pencils
V(F,Q,R,S,T...) and are said to be homographic
V (P'
when a
,
Q', R', S',
T'
...)
cross-ratio of the pencil
formed by any four lines of rays of the one
is
equal to the
corresponding cross-ratio of the pencil formed by the four corre-
sponding lines or rays of the other.
56.
Prop.
Two homographic
pencils are mutually pro-
jective.
For two
let
PQRS..., P'Q'R'S'... be any two transversals of the V and the vertices of the pencils.
pencils,
V
Let PQ"R"S"... be the common range into which these can be projected by vertices and 0'.
Then by means of a vertex K on OV the pencil (P, Q", R", S" ...); ViP,Q,R,S...) can be projected into and
this last pencil can,
by a vertex
0' (P, Q", R", S"...), that
is,
L
on 00', be projected into
0' (P', Q', R', &'...) and this
CROSS-RATIOS again by means of a vertex
M on
O'V
55 can be projected into
V'{P',Q',R\S'...).
57.
We
will
conclude this chapter with a construction for
drawing through a given point in the plane of two given parallel lines a line parallel to them, the construction heinrj effected bi/ means of the ruler only.
Let A(o,
AiOi'
be the two given
lines, &>
and
to'
being the
point at infinity upon them, at which they meet.
Let
P
be the given point
Draw any any point Join Join
B
line
upon
in the plane of these lines.
AC to cut the given lines
PA cutting Aico' in A^. PB cutting A^w' in Bi and
Join PC.
Let
in
it.
A^A and BoC meet
in Q.
Aco in
B.,.
A
and
C,
and take
CROSS-RATIOS
56
meet
GP
in 0.
Let A^O and
AC
meet
Let
PD
QB
shall
in D.
be the line required.
For
= B {A,B,Cco') = (A^BUoo) = ^i (Ao^BUo)) = {BB,PB,) = C(BB,PB,) = {AQPA,) = (AQPA,) = {ABGD). P (A,B,Cco') = P (ABCD).
{A^B.Coi')
.-.
.'.
that
is,
PD
is
PD
and
Pco' are in the
same
line,
parallel to the given lines.
EXERCISES 1.
If
{ABCD) = -^ and B be the point of C is the other point of trisection
towards A, then
trisection of of
AD
AD.
2, Given a range of three points A, B, C, find a fourth point on their line such that (ABCD) shall have a given value.
D
CROSS-RATIOS If the transversal
3.
a
{A, B, C, D),
pencil
57
,
ABC be parallel
to
OD, one of the rays
of
then
0(ABCD) = ^. {ABCD) = {ABG'D'),
4.
If
5
If A, B, C,
then each of these 6.
median
Of the
D
he
a.
range of four separate points and
{A£GD) = {AI)CB), ratios = — 1.
—
formed by the circumcentre, and orthocentre of a triangle, eight
cross-ratios of the range
point, nine-points centre
are equal to
then {ABCC') = {ABDD').
eight to
1,
2,
and eight
to
-J-.
Any plane will cut four given planes all of which meet in a common line in four lines which are concurrent, and the cross-ratio 7.
of the pencil formed 8.
Taking
A, B, C,
D
by these
a, b, c,
d
lines is constant.
to be the distances from
to the points
a line with 0, and
all in
X=(a-d){h-c),
iJ.=
shew that the six possible up of the points A, B, C,
(b- d)
(c
-
v
a),
=
(c
-
d) (a
-
b),
cross-ratios of the i-angcs that can be
D
are
/u,
I'
V
A
A
/A
J'
/x
A
I'
/i.
A
made
58
CHAPTER VI PERSPECTIVE Def.
58.
P, Q, R, S, &c.
A
figure consisting of
is
said to be in perspective with another figure
an assemblage of points
consisting of an assemblage of points P', Q', R', S', &c., lines joining corresponding points, viz.
concurrent in a point 0.
The point
if
the
PP', QQ', RR', &c. are is
called the centre
of
perspective.
It
is
clear from this definition that a figure
on to a plane or surface
is in
perspective with
when
projected
its projection,
the
vertex of projection being the centre of perspective.
seems perhaps at first sight that in introducing the notion we have arrived at nothing further than what we already had in projection. So it may be well to compare the two things, with a view to making this point clear. It
of perspective
Let plane
it
then be noticed that two figures luhich are in the same
may be
in perspective,
whereas we should not in this case
speak of one figure as the projection of the other.
In projection we have a figure on one plane or surfice and project
it
by means of a vertex of projection on
to another plane
or surface, whereas in perspective the thought of the planes or
surfaces on which the two figures
necessary
is
lie is
absent,
and
all
that
is
that the lines joining corresponding points should be
concurrent.
So then while two
figures each of
the other are in perspective,
it is
of two figures in perspective each
which
is
the projection of
not necessarily the case that is
the projection of the other.
It is clear from our definition of perspective that if 59. two ranges of points be in perspective, then the two lines of the ranges must be coplanar.
PERSPECTIVE
For
if
A, B,
C, &c. are in perspective
59 with A', B',
be the centre of perspective, A'B' and plane, viz. the plane containing the lines
It
is
But
AB
proposition will shew under what condition this
Prop.
//
to itself
perspective.
For
let
Let BB',
homographic.
{ABODE. ..) = {AB'C'B'E'.
CC
meet
The is
following
the case.
homographic ranges in the same plane of intersection of their lines is a point in the two ranges, then the ranges are in
tivo
he such that the point
corresponding
and same
&c.,
not necessarily the case that two homographic
is
ranges in the same plane are in perspective.
60.
,
OA, OB.
also clear that ranges in perspective are it
C
are in the
in 0.
.
.).
PERSPECTIVE
60
OD
Join
to cut
AB'
Then
in D".
{AB'G'D')
.•.
D' and
= {ABCB) = {AB'C'D").
B"
coincide.
(§ 51.)
any two corresponding points in the therefore they are two homographic ranges passes through
Thus the
line joining
;
in perspective.
Two pencils V(A,
61.
will according to
V are in
in perspective, points in
V'A' points ,
We If
B, Q,
D ...)
smd
V {A', B',
C',D'...)
our definition be in perspective when
in
VB
VA
V and
in perspective with points
in perspective with points in V'B'
and
can at once prove the following proposition tiuo
pencils in different planes he in pej'spective they have
a common transversal and are homographic. Let the pencils be V{A,B, C,D...) and
coplanar
(§ 59),
be
P
;
let
V (A', B',0',D'...).
VA
and V'A', which are that of VB, V'B' be Q and so on.
Let the point of intersection of
;
PERSPECTIVE
The
points F, Q, R, S, &c. each
the pencils, that
is,
they
lie in
lie
61 in both of the planes of
the line of intersection of these
planes.
Thus the points are coUinear, and
since
V(ABCD...) = (PQRS..:)^
V'iA'B'C'D'...),
the two pencils are homographic.
The
line
PQRS... containing the points
corresponding rays
According
62.
is
of intersection of
called the (uis of perspective.
to the definition of perspective
beginning of this chapter, two pencils
in
the
given at the
same plane are
always in perspective, with any point on the line joining their vertices as centre.
Let the points of intersection of corresponding rays
be, as in
the last paragraph, P, Q, R, S, &c.
We
now prove
cannot
P, Q, R, S... to be collinear, for indeed
they are not so necessarily.
But
the points P, Q,
if
Szc.
are collinear, then
we say that
the pencils are coaxal. If the pencils are coaxal they are at once seen to be honio-
graphic.
63.
It
usual with writers on this subject to define two
is
pencils as in perspective if their corresponding rays intersect in collinear points.
The
objection to this
method
is
that you have a different
definition of perspective for different purposes.
We
shall find it conducive to clearness to
definition
we have already
given,
pencils as coaxally in perspective
and we if
keep rigidly
shall
to the
speak of two
the intersections of their
corresponding rays are collinear.
As we have always coaxal
;
seen, two non-coplanar pencils in perspective are but not so two coplanar pencils.
Writers, when they speak of two pencils as in mean what we here call coaxally in perspective.' '
perspective,
PERSPECTIVE
62
Prop. If two homographic pencils in the same plane 64. have a corresponding ray the same in both, they are coaxally in perspective.
Let the pencils be
V (A, Let
and
B, G, D, &c.)
common
with the
VB
VD' in
S,
V (A,
B',
C,
D', &c.)
V'VA.
ray
and V'B' intersect and so on.
B'
in 0,
FC and V'C
in y,
VD and
B 0'
Let 7^ meet V'VA in a, and in §1 and h^ respectively.
let it
cut the rays
VD
and
VD
and
VD'
Then
since the pencils are homographic,
V(ABGD)= and
coincide with
Therefore
Sj
Thus the
intersection
VD'
lies
S..
on the line
V'(AB'C'D').
S.
of the corresponding rays
/Sj.
Similarly the intersection of any two other corresponding rays
lies
on this same
line.
Therefore the pencils are coaxally in perspective.
63
PERSPECTIVE
Prop.
65.
If ABC..., A'B'C'...,
graphic ranges not having a tiuo
he two coplanar
common corresponding
homo-
point, tlien if
pairs of corresponding points he cross-joined {e.g. AB' and all the points of intersection .so obtained are collinear.
A'B)
Let the lines of the ranges intersect in P.
Now
according to our hypothesis
P
is
not a corresponding
point in the two ranges.
be convenient to denote
It will
and
Y',
to the
according as
A'B'C ...
we
consider
it
P by two
different letters,
to belong to the
X
ABC... or
range.
Let X' be the point of the A'B'G'... range corresponding to Y be the point of the ABC... range
A" in the other, and let
corresponding to
Then
]"'
in the other.
{ABCX Y...) = [A'B'C X'Y'. .-.A' {ABCXY. ..) = A (A'B'CX'Y'...). ).
.
.
These two
})encils
have a
common
ray, viz. Axi', therefore
by
the last proposition the intersections of their corresponding rays are collinear,
viz.
A'B, AB'; A'C,
and so
From
this
it
will
of the cross-joins of
on
is
AC;
A'X, AX'; A'Y,AY';
on.
the line X'Y.
be seen that the locus of the intersections
A
and A' with
B and
B',
C and
C
and
so
64
PERSPECTIVE Similarly the cross-joins of any two pairs of corresponding
points will
lie
on A'T.
X'Y
This line
called the liomogntpJiic axis of the
is
two
ranges.
This propositi(jn
is
also true if the
The proof
corresponding point.
two ranges have a
may
of this
be
common
left
to the
student. 66.
The student may obtain
this chapter
by proving that
V {A,
B,
C.)
practice in the methods of
if
V {A', B'
and
,
G'...)
be two homographic coplanar pencils not having a common corresponding ray, then if we take the intersections of VF and F' {VF, V'F'; and VQ, V'Q' being Vq, and of VQ and
V
any two pairs of corresponding
lines)
and join
these, all the
lines thus obtained are concurrent.
It will
be seen when we come to Reciprocation that this
proposition follows at once from that of § Qb.
TRIANGLES IN PERSPECTIVE 67.
Prop
// the vertices of two
.
triangles are in perspective,
the intersections of their corresponding sides are collinear,
and
conversely. (1)
Let
Let the triangles be in different planes. be the centre of perspective of the triangles
ABC,
A'B'C. Since BC, B'C are in a plane, viz. the plane containing OB and OC, they will meet. Let A' be their point of intersection. Similarly
GA
and G'A'
will
meet
(in
Y) and
AB and
A'B'
(in Z).
Now
A'',
V,
Z are
in the planes of both the triangles
ABG,
A'B'G'.
Therefore they
lie
on the
line
of
intersection
planes.
Thus the
first
part of our proposition
is
proved.
of these
65
PERSPECTIVE
Next
let
the triangles
ABC, A'B'C be such
that the inter-
sections of corresponding sides {X, Y, Z) are collinear.
BC and
Since
EC meet they are coplanar, and similarly for
the other pairs of sides.
Thus we have three planes BCOB', CAA'C, which AA', BB', CO' are the lines of intersection. But three planes meet Therefore AA', BB',
ABB' A',
of
in a point.
CC
are concurrent, that
is,
the triangles
are in perspective. (2)
Let the triangles be in the same plane.
First let
Let
them be
A", Y,
Z be
in perspective, centre 0.
the intersections of the corresponding sides
as before.
Project the figure so that A'^Fis projected to infinity. A. G.
6
PERSPECTIVE
QG
Denote the projections of the different points by corresponding small letters.
We
have now oh
that
oh'
:
let
parallel to
oc
oa' since ca is parallel to c'a'.
.".
ah
.'.
z
is
parallel to
is
h'c'
n'b'.
at infinity also,
is
A',
.*.
Next
since be
:
:
are collinear.
X, y, z
is,
= oc = oa
F,
A^,
Z
F,
Z are
collinear.
be collinear;
we
prove that the
will
triangles are in perspective.
Let A A' and BB' meet
OC
Join
Then
and
let it
A'C
ABC and A'B'C" are
the intersection of
.-.
But
in 0.
meet
BG and .•.
BG
in G".
in perspective.
and B'G"
B'G' meet the line
B'G" and i.e.
Thus
ABG and
68.
Prop.
G"
lies
YZ in
i?'C" are in the
coincides with
on the
line
YZ.
X by hypothesis.
same
line,
C".
A' B'G' are in perspective.
The necessary and sufficient condition that ABG, A' B'G' should he in perspective is
coplanar triangles
AB,
.
AB., .GA,.
GA^ BC, .
.
BG.,
= AG,.AG,.BA,. BA,
.
GB, GB,, .
the
PERSPECTIVE A^,
A2 being
the points in
67
which A'B' and
A'C
meet the
non-corresponding side BC,
B„
Bo being the points in ivhich
B'C and B'A'
meet the non-
con-esponding side CA, Ci, Co being the
points in which
corresponding side
CA' and
C'B' meet the non-
AB.
First let the triangles be in perspective; let
XYZ be
axis of perspective.
5—2
the
PERSPECTIVE
68
Then
X, B^
since
Co are collinear,
,
A B, CX BO, AC.BX.CB, .
•'•
.
Since Y, C\ Ao are collinear, ,
AY.CA,.BC\ AC,.BA.,.CY
*
=
1.
Since Z, Ai, Bo are collinear,
AB^.CA,.BZ _
'az.ba,.cb~ Taking the product of these we have
AB, AC,
.
AB.,
.
.
AG.,
.
But X,
Y,
GA, GA,. BC, BC,. .
BA,
Z are
.
BAo^
.
C%
.
CB.,
.
AY.GX BZ AZ BX CY~ .
.
.
collinear,
•
.-.
.
•
AY.GX.BZ AZ.BX.CY~
AB,.AB,.GA,.CA,.BC\.BC., =^AG,.AC,.BA,. BA, CB, GB,. .
Next we can For
it
sheAv that this condition
.
is sufficient.
renders necessary that
AY.CX .BZ AZ.BX.GY' .".
X,Y,Z are
Cor.
collinear
If the triangle
and the triangles are in perspective.
ABC
and the points A,, Ao, B,, Bo, proposition, also
it is
be in perspective with A'B'C', Co be as defined in the above
Cj,
clear that the three following triangles
be in perspective with
ABC,
must
viz.
(1) the triangle
formed by the lines A^Bo, B^C,, CoA^,
(2)
„
„
„
„
(3)
„
„
„
„
A,B„ A,B„
B,Co_,
G,A„
B,C\, C,A,.
PERSPECTIVE
69
EXERCISES 1. ABC, A'B'C' are two ranges of three points in the same plane; i?C" and 5'C intersect in ^i, CA' and A in B^, and AB'
C
and A'B
in Cj
;
prove that A^, B^, Cj are collinear.
ABC
2.
and A'B'C are two coplanar triangles in perspective, any line is drawn not in the plane of the and .S" are any two points on this line. Prove that the
centre 0, through triangle
means
^S*
;
ABC
triangle
by means
of tlie centre S,
and the triangle A'B'C by
of the centre S', are in perspective
with a
common
triangle.
Assuming tliat two non-coplanar triangles in perspective are by means of Ex. 2 that two coplanar triangles in per-
.3.
coaxal, prove
spective are coaxal also.
ABC, A'B'C
If
4.
and B'C
be two triangles in perspective, and
intersect in A^,
CA' and
CA
in B^,
AB' and A'B
if
BC
in Cj,
then the triangle A^B^C^ will be in perspective with each of the given triangles, and the three triangles will have a
common
axis of
perspective.
When three triangles are in perspective two by two and 5. have the same axis of perspective, their three centres of perspective are collinear.
on the straight line AC, and the VQ meets the straight line AB in is anotlier point on AB XQ meets meets AB in Y in U, and XR meets AD in II', pro\e that YU, ZW, JC are
The points Q and
6.
Ton and VR
point Z,
AD
Ji lie
the straight \ine_AD
;
X
;
;
concurrent.
ABC,
A! B'C should be in perspective
AV.Bc' where a, A' B' ,
to
sutHcient condition that the coplanar
The necessary and
7.
triangles
,
b',
c
.
Ca =
Ac'
.
Ba
.
is
Cb',
denote the sides of the triangle
A'B'C
opposite to
C respectively, and Ab' denotes the perpendicular from A
on
b'.
X
CA' and CA in Y A'B' and B'C and BC meet in The condition given ensures that X, Y, Z are collinear.]
[Let in Z.
;
;
AB
PERSPECTIVE
70
Prove that the necessary and
8.
coplanar triangles
ABC, A'B'C should be
ABC sin ABA' sin BCA' sin BC B' sin CAB' sin CAC _
sin
A CB'
[This
A CA' sin CBA'
sin
sin
CBC
BA C
sin
Let him turn to
command
is it difficult
BAB' ~
'
to
remember
if
The
result
is
aX A',
easily obtained.
the student grasps the principle,
mind
—the
principle, that
round the triangle in the two opposite directions, (2) AC, GB, BA.'\
Two
The
to establish the test
these formulae relating to jjoints on the sides of a
all
triangle are best kept in
9.
sin
'
36 Cor., and take in turn
§
C and at the centre of perspective.
by which
'
proved in Lachlan's Modern Pure Geometry.
is
for himself.
Nor
in perspective is
sin
student has enough resources at his B',
,
sufficient condition that the
triangles in plane perspective can
is,
(1)
of travelling
AB, BC, CA,
be projected into
equilateral triangles.
ABC
10.
meets
is
a triangle,
IJ^ meets
respectively.
AB in
/j, /o, L^ its
BC
in A^,
Ci, prove that A^, B^,
ecentres opposite to A, B,
/.j/j
C'j
meets
CA
in B^
and
C
/j/,
are collinear.
CF and AD', BE', CF' be two sets of condrawn through the vertices of a triangle AP>C and meeting the opposite sides in D, E, F and D' E', F', and if EF and E' in Z, E'F' intersect in X, FD and F'D' in 7, and DE and 11.
\i
AD, BE,
current lines
,
U
then the triangle
XYZ is
in perspective
with each of the triangles
ABC, DEF, D'E'F'. [Project the triangle so
that
pendiculars in the projection and
then use Ex.
7.1
AD, BE, CF become the perAD' BE', CF' the medians, and ,
71
CHAPTER
VII
HARMONIC SECTION Def.
69.
Four collinear points A, B,
harmonic range
C,
D
are said to form
if
{ABCB)=-l.
We
have in this case
AB.CD =
AJJ
.
-1.
CB
AB-AC^ AB-A G AD~ AD-AC AG- AD'
AB ''
thus
AC is
Now
a harmonic
a range of four points all
mean between
AB and
AD.
reverting to the table of the twenty-four cross-ratios of
we
(§ 50),
the follow ing cross-ratios
=—
see that
if
(A BCD)
= — 1,
then '
1
:
"^^
\aBCD\ (BADC), (GDAB), (DCBA), (ADGB), (BCDA), (CBAD), (DABC). Hence not only
AD, but
BD DB CA
We B
is
AC
a harmonic
mean between
AB
and
also
shall
is
a harmonic „
mean between „
„
„
„
then speak of
A
and D, and express the
BA DC CB
„
and
C
and BC, and DA, and CD.
as harmonic conjugates to
fact symbolically
(AC,BD) = -1.
thus
HARMONIC SECTION
72
By and and
we mean
this
in which, it will
B
and
that
all
the eight cross-ratios given above,
A
be observed,
D alternate,
C are alternate members,
and
are equal to
—
1.
D as the fourth AG is divided BD is so divided at A and C.
When (AC, BD) = - 1 we sometimes speak of harmonic of J., i? and C or again we say that ;
B
harmonically at
and D, and that
Or again we may say with respect to
A
pencil
P
B
G
that
is
harmonically conjugate with
A
and D.
{A,B,
G,
D)
of four rays
the points of intersection of
is
called
harmonic when
rays with a transversal form a
its
harmonic range.
The student can
easily prove for himself that the internal
and external bisectors of any angle form with the it a harmonic pencil.
Prop.
70.
// {AG, BD) = -1, and
lines containing
be the middle
jmnt
of AG, then
OB.OB=OG' = OA\ B
{ABGD) = -
For since .-.
1,
AB.GD = -AD.GB.
Insert the origin 0.
(OB- OA)(OJJ- OG) = -(OD - OA)(OB -OG).
.-.
0A=- OG.
But .•; .-.
(OB+OC)(OD-OG)=-(OD+OG)iOB-OG). OB.OD + OG.OD-OB.OG- OG"= - OD OB + OG OD - OB OG + 0G\ 20B.On = 20G'. 0B.0D = 0G"- = A0'' = 0A\ .
.
.
.-.
.-.
Similarly
if 0'
Cor.
The converse
be the middle point of BD,
0'G.O'A = 0'B'=0'D\ that
if
OG' =
1.
ABGD
OB OD, .
of the above proposition
be a range and then (AG,
is
the middle point of
BD) = -
1.
This follows by working the algebra backwards.
true, viz.
^Cand
73
HARMONIC SECTION Cor.
Given three points A, B, C that (AB, CD)
2.
point
D
in the line such
circle
on
AB as
diameter, then
// {A C, BD) =
Prop.
71.
D
is
—
1,
Let
be the middle point of
Let this
we
1
describe a
the inverse point of C.
A C as dia meter
the circle on
will cut orthogonally evenj ciixle through
of the circle on
in a line, to find a
=-
AC
B and
D.
and therefore the centre
A C.
circle cut
any
circle
through
B
and
D
in
P
;
then
OB.OD=OC'=OP\ Therefore
OP
is
a tangent to the circle
BrD\
thus the
circles cut orthogonal 1}-.
Similarly, of course, the circle on
A and C. ABCD he a range,
BD
will
cut orthogonally
every circle through
Cor.
1.
If
and if
the circle
on
diameter cut orthogonally some one circle passing through
D,then
AC B
as
and
(AC,BD) = -1.
For using the same figure as
before,
we have
OB.OD = OP'=:OC\ .-.
Cor. is
2.
If fioo
{AC,BD) = -1.
ciixles cut orthogonally,
divided harmo)iically by the other.
any diameter of one
HARMONIC SECTION
74
Prop.
72.
angle, then
PC
PA
If P(AB, CD) and PB are the
= -1 and bisectors
APB
be
a right
of the angles betiueen
and PP.
Let any transversal cut the rays PA, PB, PC, harmonic pencil in A, B, C, D.
PD
of the
Tht
AD.BC AC:AD=CB:BD. P lies on the circle on AB as diameter we have by PC:PD = CB:BD=AG:AR PA and PB are the bisectors of the angle GPD. •
•
.-.
as
.•.
§27
.".
73.
Prop.
If on a chord
PQ
of a circle two conjugate
points A, A' with respect to the circle be taken, then
{PQ,AA') = ~l.
Draw which A' Let
the diameter lies,
CD
through
A
to cut the polar o^
A, on
in L.
be the centre.
Then by the property
of the polar,
OL.OA=^ .-.
Therefore the circle on will cut orthogonally
But the
circle
on
0C-\
{CD,LA) = -1.
CD
as diameter
every circle through
A A'
(i.e.
A
the given circle)
and
L
(§ 71).
as diameter passes through
A
and L.
Therefore the given circle cuts orthogonally the circle on
A A'
as diameter.
HARMONIC SECTION But the given
circle passes .'.
P
through
usefulness.
It
Chords of a divided at the
A
may be
ciixle
and
and
Q.
(PQ,AA') = -1.
This harmonic property of the circle
and
75
of great importance
is
otherwise stated thus
through
A
a point
:
are harmonically
at the point of intersection of the chord with
polar of A.
Prop.
74.
lateral
is
Each of
Let AB, BC, CD, A, B, C, D, E, lines
taken in
F its
of a plane quadn-
PQR
Project
DA
be the four lines of the quadrilateral
six vertices, that
is,
the intersections of
its
pairs.
Then AC, BD, Let
the three diagonals
divided harmonically by the other two.
EF ave
be the
EF to
its
ti-iangle
infinity.
by corresponding small
diagonals.
formed by
its
diagonals.
Denote the points
in the projection
letters.
bp
Then
bq
.
.
dq
_
bp
dp
dp
1, bci
= -1.
q being at
x
HARMONIC SECTION
76
{AFCR) = -l. (FQER = n {FQER) = (.4 FOR) = -1.
Similarly
Also
)
Thus we have proved
{AC,FR) = -l, The
Cor.
(BD,FQ) = -1,
FQR
circumcircle of
{EF,QR) = -l.
will cut orthogonally the
three circles described on the three diagonals as diameters.
Note. that
the
if
It has
been incidentally shewn in the above proof
M be the middle point of AB, a the point
75.
The harmonic property
the last article,
on
at infinity
= -l.
line, (/i^, il/a))
of the quadrilateral, proved in
of very great importance.
is
It
is
important
too that at this stage of the subject the student should learn "to
take the 'descriptive' view of the quadrilateral; for in 'descriptive geometry,' the quadrilateral
containing an area plane, which
meet
;
is
not thought of as a closed figure
but as an assemblage of four
a
lines in
in pairs in six points called the vertices
;
and
the three lines joining such of the vertices as are not already
joined by the lines of the quadrilateral are called diagonals. opposite vertices
we mean two
By
that are not joined by a line of
the quadrilateral. 76.
A
A (|uadrilateral is to be distinguished from a quadrangli
quadrangle
is
to
I
be thought of as an assemblage of fou
points in a plane which can be joined in pairs
by
six straight
HARMONIC SECTION lines, called its sides or lines
two of these
which do not
sides
in a point of the quadrangle are called opposite sides.
meet
And
;
77
the intersection of two opposite sides
point.
This
name
is
is
called a diagonal
not altogether a good one, but
it is
suggested
by the analogy of the quadrilateral. Let us illustrate the leading features of a quadrangle by the accompanying figure.
ABCD is the quadrangle. Its sides are AB, BC, CD, DA, AC and BD. AB and CD, AG and BD, AD and BC are pairs of opposite sides and the points P, Q, R where these intersect are the diagonal points.
The
triangle
PQR
may be
The harmonic property
called the diagonal triangle.
of the quadrangle
is
that the two
sides of the diagonal triangle at each diagonal point are
conjugates
IV ith respect to the
harmonic
two sides of the quadrangle meeting
in that point.
The student will have n(^ difficulty in .seeing that this can be deduced from the harmonic property of the quadrilateral proved in § 74,
On
account of the harmonic property, the diagonal triangle been called the harmonic
associated with a quadrangle has triangle.
HARMONIC SECTION
78
EXERCISES If
1
that
it is
M
and
JS^
be points in two coplanar lines AB, CJ), shew and N' project into the middle
possible to project so that
points of the projections of
AB
and
M
CJJ.
AA^, BB^, CC-i are concurrent lines through the vertices of a B^C^ meets BG triangle meeting the opposite sides in yli, B^, 6\. A^B^ meets AB in Cgj prove that in A.,; C, .Ij meets CA in 2.
Z>._,
{BC, A,A.^ = -
;
= -l,
{CA, B,B,)
1,
{AB, C,C.^^-l.
Prove that the circles described on the (as defined in Ex. 2) as diameters are coaxal.
lines A^A.,, B^B.^, C\ Co
3.
[Take
P
a point of intersection of circles on
shew that CjCo subtends a right angle at
The
4.
collinear points A, D,
fixed line through C,
E
is
From any
5.
MC
and meeting intersect
M'B'
:
in
point
intersect in Q,
AC
in P.
M in
AB P,
and
AP
and
§ 27.]
is
Prove that
the side
drawn parallel and AC in B' and
and
CB is any other any moving point the lines CQ and DE
B
BD
are
2
are given:
a fixed point, and
on CE. The lines A E and in R, and the lines BR and as B moves along G E.
MB' and
C
A-^Ac^, B^B,,,
Use Ex.
F.
P is
a fixed point
BC of a triangle ABC lines AC and AB respectively,
to
intersects
C".
The
B'C
in
lines
BC
M'.
and GB' Prove that
M'C'^MB MC. :
harmonic conjugates {DD'), {EE'), {EF') are respectively taken on the sides BC, CA, AB of a triangle ABC with respect to the pairs of points {BC), {CA), {AB). Prove that the corresponding sides of the triangles DEE and D'E'E' intersect on the sides of the triangle ABC, namely EE a.nd E'F' on BC, and Pairs
6.
of
so on.
The -lines
7.
VB', VC bisect the internal angles formed any point V to the angular points of the triangle on BC, B' on CA, G' on AB. Also A", B", C"
VA',
by the
lines joining
ABC
and A'
;
lies
are harmonic conjugates of A',
and A,
A and
B', 6"
with respect to
Prove that A", B", C" are
B.
B
and
G,
C
collinear.
BB^, CC^ are the perpendiculars of a triangle ABC in Cg; is the middle point of line joining A to C'jX and BB^ meet in 7\ Prove that C.2T is the orthocentre perpendicular to BC. 8.
A^Bj^
AA-^^,
meets
X
AB
;
;
HARMONIC SECTION
^
9.
i^
the tangent at P.
^
the locus of
A
10.
and P a variable ^i^ at right angles to AP the rectangle FAPQ be completed
a fixed point without a given circle
is
point on the circumference.
meets in
79
is
a straight
line is
The If
line
line.
drawn cutting two
non-intersecting circles; find
a construction determining two points on this line such that each is the point of intersection of the polars of the other point with respect
two
to the 11.
circles.
A^, B^, C\ are points on the sides of a triangle
to A, B, C.
A^, B.,,
0-2.
harmonic conjugates with
A and B. If CCi be concurrent.
with
ABC opposite
are points on the sides such that Jj,
A.,, B.^,
B
and
C
B^, Bo with
;
C and A
;
A.^
are
C\, C^
C, are collinear, then must AA^, BB^,
12. AA^, BBy, CCi 3-re concurrent lines through the vertices of a triangle ABC B^C^ meets BC in A.^, C,Ji meets CA in B.,, A^B^ meets AB in Cj. Prove that the circles on A^A^, B^B.^, C^C^ as diameters all cut the circumcircle of ABC orthogonally, and have
their centres in the
[Compare Ex.
same straight
line.
3.]
If .1 and B be conjugate points of a circle and J/ the 13. middle point of AB, the tangents from to the circle are of
M
MA.
length 14.
If
a system of circles have the same pair of points con-
jugate for each circle of the system, then the radical axes of the circles,
15.
taken in If a
pairs, are concurrent.
system of
circles
have a
common
pair of inverse points
the system must be a coaxal one.
and 0' are the limiting points of a s3'stem of coaxal and A is any point in their plane shew that the chord of contact of tangents drawn from A to any one of the circles will 16.
circles,
;
pass through the other extremity of the diameter through circle
AGO'.
A
of the
80
CHAPTER Yin INVOLUTION Definition.
77.
be a point on a line on which
If
B, Bi
;
C,
C'l
;
lie
pairs of points A, A-^
;
&c. such that
= k,
OA.OA, = OB.OB, = OC.OC,= the pairs of points are said to be in Involution. points, such as
A
and A^, are
each of two conjugates
The point
is
is
called conjugates;
associated
and sometimes
called the 'mate' of the other.
called the Centre of the involution.
If k, the constant of the involution,
conjugate points
Two
lie
be positive, then two
on the same side of 0, and there
will
be
K, K' on the line on opposite sides of that is such that each is its own mate in the involution and K' are called the double OK^ = OK'- = k. These points
two
real
points
;
K
points of the involution. It is
that
is
K
important to observe that
why we
write K' and not
It is clear that
{AA^, KK') =
is
not the mate of
K'
;
iLj.
—
1,
and so
for all the pairs of
points.
two conjugate points will lie on opposite and the double points are now imaginary.
If ^ be negative,
sides of 0,
If circles be described on AA-^, BB^, CCi, &c. as diameters
they will form a coaxal system, whose axis cuts the line on
which the points
lie in 0.
INVOLUTION
,
81
K and K' are the limiting points of this coaxal system. Note
also that for every pair of points, each point is inverse
to the other with respect to the circle
on
KK'
as diameter.
an involution is completely determined when two pairs of points are known, or, what is equivalent, one pair of points and one double point, or the two double points. It is clear that
We
must now proceed to establish the criterion that three the same line may belong to the same
pairs of points on involution.
The necessary and
Prop.
78.
pair of points C, C'l should belong by A,\-i,; B, B, is
condition that a
siifficient
to the involution
determined
{ABCA,)={A,B,C\A). First
its
we
will
shew that
this cijnditinn is necessary.
Suppose C and C\ do belong centre and k its constant. .
•.
OA
OA, =
.
/
OB
k
k
.
\
to the involution.
OH,
/
= OC 0C\ = .
k
k
Let
be
k.
\
[ob,~oaJ \oa~ocJ [oA OaJ \0n\ ~ 00 OB,OA,) {OA -0C\) ^ A,B,.C,A _ ^ {OA- OA,) {OB, - OC,) 'A,A C\B, (
y^^^^^'^^^)-
.
Thus the condition
is
necessary.
[A more purely geometrical proof of
theorem
will
be
in the involution determined
by
this
given in the next paragraph.]
Next the above condition For
and
let
let
C be
is sufficient.
{ABCA,) = (A,B,C,A) the mate of
C
A,A,i B,B,. A. G.
6
INVOLUTION
«2
..{A,B,C\A) = {A,B,G'A). .*.
Cj
Hence the proposition Cor.
1.
If A, A,;
and 6" is
coincide.
established.
B,B,; C,C,; D, D, belong
to the
same
involution
{ABGI)) Cor.
2.
If
= {AACM.
K, K' be the double points of the involution
{AA.KK') = {A.AKK') and (ABKA,) = {A,B,KA).
We may
79.
prove the
first
part of the above theorem as
follows.
If the three pairs of points belong to the
same
involution,
the circles on AA^^, BB^, CCi as diameters will be coaxal
Let
P be a point
Then
of intersection of these circles.
AFA,, BPB,, CPC\
the angles
(§ 77).
are right angles
and
therefore
P(ABCA,) = F(A,B,C\A). (ABCA,) = (A,B,C,A).
.-.
The still
circles
may
not cut in real points.
But the proposition
holds on the principle of continuity adopted from Analysis. 80.
The
proposition
greatest importance.
we have
just proved
is
of the very
INVOLUTIOX The
criterion that thr^ee pairs
83
of points belong
the
to
same
involution is that a cross-ratio formed with three of the points,
one taken
from
should be equal
each pair, and the mate of any one of the three to the corresponding cross-ratio formed by the
mates of these four points.
matter in what order we
It does not of course letters
wi'ite
provided that they correspond in the cross-ratios.
the
We
could have had
{AA,CB) = {A,AC,B,) {AA,C\B) = {A,ACBy).
or
All that
essential
is
cross-ratio, three
A
Prop.
81.
is
that of the four letters used in the
should form one letter of each pair.
range of points in involution projects info a
range in involution.
For
let
A, A^; B, B^; C, C\ be an involution and
projections be denoted
by corresponding small
{ABCA ,) = {A, B, C, A {ABC A,) = {abca,) (zi,B,C^A) = {aAciCi). {abca^) = {a^biCia).
Then But and
let
the
letters.
).
.'.
.•.
Note.
a,
The
tti
:
6, 61
;
r,
d form an
involution.
centre of an involution does not project into
the centre of the involution obtained by projection; but the
double points do project into double points. 82.
We
Involution Pencil.
now
see that
VP,
if
we have
VF;
a pencil consisting of pairs of
VQ, VQ'; VU, VR' &c.
such that any transversal cuts these in pairs of points
A, A,
;
B,B,\
C, C, &c.
forming an involution, then every transversal
will cut the pencil
so.
Such a pencil will be called a Pencil in Involution or simply an Involution Pencil.
6—2
84
V
INVOLUTION
The double lines of the involution pencil are the lines through on which the double points of the involutions formed by
different transversals
lie.
Note that the double any pair of conjugate
From
lines are
harmonic conjugates with
rays.
this tact it results that if
VD
and
VU
he the double
of an involution to ivhich VA, VA^ belong, then VD and VD' are a pair of conjugate lines for the involution wJtose double
lines
VA, VA^.
lines are
We
83. to
shall
postpone until a later chapter, when we come
deal with Reciprocation, the involution properties of the
quadrangle and quadrilateral, and pass now to those of the circle which are of great importance. We shall make considerable
them when we come
use of
Involution properties of the circle.
84.
Prop. a
line
Pairs of points conjugate for a
form a range
the points
Let Let
circle
lie
line
with the
OPQ
is
CK be
a self-conjugate triangle, and
the perpendicular from
C
on
PK.KQ= KO.KC. .-.
line
I.
/.
at G, the centre of the circle (§ 16 a).
Then
along
circle.
and Q be a pair of conjugate points on the
be the pole of
Let
which
in involution of ivhich the double points are
of intersection of the
P
Thus is
to treat of the Conic Sections.
KP.KQ=OK.KC.
I.
its
orthocentre
INVOLUTION
85
Thus P and Q belong to an involution whose centre is K, and whose constant is OK KG. Its double points are thus .
or imaginary according as
real
PQ
does or does not cut the
circle.
PQ cut the circle in A and B, then OK KG = KA- = KB\ see that A and B are the double points of the involu-
If
thus
.
we
tion.
A
It is obvious too that
since each
The
is its
own
and
B
must be the double points
conjugate.
following proposition, which
foregoing,
easily
is
Prop.
deduced from
Pairs of conjugate
lines
form an involution tangents from the point.
through a point lines are the
is
the reciprocal of the
it.
for a
circle,
luhich
For pairs of conjugate lines through a point the polar of
in
pairs of
pass
pencil of which the double
will
meet
conjugate points, which form an
involution range, the double points of which are the points in
which the polar of
Hence the
cuts the circle.
pairs of conjugate
lines
through
form an
involution pencil, the double lines of which are the lines joining to the points in
which
its
polar cuts the circle, that
is
the
tangents from 0.
be within the
If 85.
A
circle the
double lines are not
real.
Orthogonal pencil in Involution.
special case of
an involution puneil
is
that in which each
of the pairs of lines contains a right angle.
That such a pencil theorem of
is
§ 84, for pairs
in involution
is
clear from the second
of lines at right angles at a point are
conjugate diameters for any circle having
its
centre at that
point.
VP, VP^; by taking any transversal t to A, A^; B, B^ &c. and drawing the perpendicular
But we can ^^Q,
cut these in
VO
also see that pairs of orthogonal lines
V^Qi &c. are in involution,
on to
t
;
then
0A.0A, = -0V'=0B.0B,.
INVOLUTION
86
Thus the
of points
pairs
belong to an invohition with
imaginary double points.
Hence
pairs of orthogonal lines at a point form a pencil in
involution with imaginary double lines.
Such an involution Note that
is
called
this property
an orthogonal involution.
may
give us a test whether three
pairs of lines through a point form
an involution.
If they can
be projected so that the angles contained by each pair become
must be
right angles, they
in involution.
Prop. In every involution pencil there is one pair of 86. rays mutually at right angles, nor can there he more than one such pair unless the involution pencil he
Let
P be
an orthogonal
one.
the vertex of the pencil in involution.
Take any
transversal
I,
and
involution range which the pencil
let
be the centre of the
makes on
it,
and
let
k be the
constant of this involution.
Join OP, and take a point P' in
Thus
P
and P'
according as Bisect
meet the
is
/..•
PP' line
I
will
in
A
such that
OP
.
OP' = k.
positive or negative.
in
M and
draw J/C
at right angles to
PP'
to
in C.
Describe a circle with centre I
OP
be on the same or opposite sides of
C and
radius
OP
or
CP'
to cut
and A^.
The
points
^1
and A^ are mates
in the involution
OA .OA,= OP.OP' = k.
on
I
for
INVOLUTION Also the angle
APA^
87
being in a semicircle
Hence the involution pencil has the
is
a right angle.
pair of rays
PA
and PA^
mutually at right angles.
In the special case where
C
and
In the case where 2T,
is
the middle point of PP',
coincide.
PO
is
perpendicular to
and the point C is at line through P parallel L,
PP'
I
and the point at
rays, for
involution range on
In this case
infinity.
to
the line through
/,
the middle point of PP', perpendicular to
parallel to
is
it is
PO
and the
which are the pair of orthogonal infinity
along
I
are mates of the
/.
Thus every involution pencil has one pair of orthogonal rays. more than one pair of rays at right angles, then all the pairs must be at right angles, since two pairs of
If the pencil have
rays completely determine an involution pencil.
Prop. An invohdion pencil projects into a pencil in imd ani/ involution can be projected into an orthogonal
87.
involution,
involution.
For
p
let
A.,
A^; B, Bi]
Then .-.
in the
the pairs of rays of an involution pencil at
plane meet the line of intersection of the
.4,
A^
C, C^ &c. ;
and
let 0'
B, B^; C, Cj &c.
we may choose two
planes in 0.
an involution range. is
an involution.
Again, as we can project two angles in the right angles,
ir
be the projection of
is
0' (A, A^; B, B^; C, C\ &c.)
p and
p
plane into
angles between two pairs of
INVOLUTION
88
rays of an involution to be so projected. projection It
must be an orthogonal
may be remarked
Then the
pencil in the
one.
too that at
tlie
same time that we
project the involution pencil into an orthogonal one
project any line to infinity
Note.
xA.s
lines, it is clear
an orthogonal involution has no that
if
we can
(§ 43).
an involution pencil
is
double
real
to be projected
into an orthogonal one, then the pencil thus projected should
not have real double
An
involution
lines, if
the projection
is
to
be a real one.
pencil with real double lines can only be
projected into an orthogonal
one by means of an imaginary
vertex of projection.
The reader
will
understand by comparing what
stated with § 43 that the two circles determining article
do not
V
is
here
in that
intersect- in real points, if the double points of the
involution range which
the involution pencil in the
p
plane
intercepts on the vanishing line be real, as they are if the
involution pencil have real double lines.
EXERCISES 1.
Any
of points
transversal is cut by a system of coaxal circles in pairs which are in involution, and tlie double points of the
involution are concyclic with the limiting points of the system of circles. 2.
If
A",
K' be the double points of an involution to which then A, B^; A^, B K, A" are in involution.
A, A^; B, By belong 3.
;
;
If the double lines of a pencil
angles, they
must be the bisectors
in involution be at right
of the angles
between each pair
of conjugate rays. 4. A.'
B'C
The corresponding
sides
BC, B'
C
ifcc.
of
in plane perspective intersect in P, Q,
AA', BB',
CC
respectively intersect the line
two
triangles
Prove that the range {PP' QQ\ BR') forms an involution. ,
ABC,
R respectively; and PQR in P\ Q', K'.
89
EXERCISES 5.
The centre
system of 6.
circles
Shew
quadrilateral also
;
lies
on the three diagonals.
that is
formed by the on the radical axis of the
of the circumcircle of the triangle
three diagonals of a quadrilateral
if
each of two pairs of opposite vertices of a circle, the third pair is
conjugate with regard to a
and that the
circle is
one of a coaxal system of which the line
of collinearity of the middle points of the diagonals
is
the radical
axis.
The two pairs of tangents drawn from a point to two circles, 7. and the two lines joining the point to their centres of similitude, form an involution. 8. Prove that there are two points in the plane of a given triangle such that the distances of each from the vertices of the
triangle are in a given ratio.
Prove also that the
line joining these
points passes through the circumcentre of the triangle.
90
CHAPTER IX THE CONIC SECTIONS Definitions.
88.
The Conic
Sections (or Conies, as they
are frequently called) are the curves of conical, or vertical, pro-
They
jection of a circle on to a plane other than its own.
then the plane sections of a cone having a circular base.
are
It is
not necessary that the cone should be a right circular one, that is,
that
its
vertex should
on the
lie
it.
has a circular base (and consequently too to the base are circles), the sections of
The
89.
through the centre of
line
the circular base and at right angles to
it
conic sections are classified according to the relation
touches the
parabola; curve
is
if
circle,
and
curve of projection
is
In other words a parabola
By
a
'
is
the vanishing line does not meet the
by a plane
circular base, cone.
If the vanishing
the curve of projection
called an ellipse;
circle the
sections parallel
are called conic sections.
of the vanishing line to the projected circle. line
So long as the cone
all its
if
called
circle,
the vanishing line cuts the
called a hyperbola.
is
the section of a cone, having a
parallel to a generating line of the
generating line
'
is
meant a
line joining the vertex
of the cone to a point on the circumference of the circle
forms
a
the
which
its base.
An
ellipse is
a section of the cone by a plane such that the
plane parallel to
it
through the vertex cuts the plane of the
base in a line external to
A hyperbola, is parallel plane
it.
a section of the cone by a plane such that the
through the vertex cuts the base of the cone.
THE CONIC SECTIONS
The curves
are illustrated
91
by the following
figure
and
it
should be observed that the hyperbola consists of two branches, and that to obtain both these branches the cone must be prolonged on both sides of
90.
Focus and
Every conic
we
its vertex.
directrix property.
section, or projection of a circle, possesses, as
shall presently shew, this property,
namely that
it is
the locus
of a point in a plane such that its distance from a fixed point in
the plane bears to
its
plane, a constant ratio.
distance from a fixed line, also in the
The
fixed point
is
called the focus of
the conic, the fixed line
is
called the directrix,
ratio the eccentricity.
It
Avill
tricity is unity, less
and the constant
be proved later that the eccen-
than unity, or greater than unity, according
as the conic is a parabola, an ellipse, or a hyperbola.
THE CONIC SECTIONS
92
Text books on Geometrical Conic Sections usually take
91.
the focus and directrix property of the curves as the definition of them, and develop their properties therefrom, ignoring for
the purpose of this development the
when
by
fact
every conic
that
and directrix property, This is to be is all the while the projection of some circle. For many of the properties which can only be regretted. evolved with great labour from the focus and directrix property
section, even
defined
are proved with great ease projections of a circle.
easy
it
is
its
focus
when the
We
conies are regarded as the
shall in the next chapter
shew how
that plane curves having the focus and
to prove
directrix property are the projections of a circle. 92.
Projective properties.
The
conic sections, being the projections of a circle
possess
all
the projective properties of the
must
circle.
They will be such that no straight line in their plane (1) can meet them in more than two points, and from points which are the projections of such points in the plane of the circle as lie without the circle, two and only two tangents can be drawn, which
will
be the projection of the tangents to the
The
(2)
property
'
conic sections will clearly have the
That
of the circle.
is,
'
circle.
pole and polar
the locus of the intersections
of tangents at the extremities of chords through a given point will be a line, the point and line being called in relation to The polar of a point from which one another pole and polar. tangents can be drawn to the curve will be the same as the line
This line is through the points of contact of the tangents. the chord of contact,' but strictly speaking the often called '
chord
The
is
only that portion of the line intercepted by the curve.
polar
(3)
is
unlimited in length.
If the polar of a point
then the polar of
B
A
for a conic
must go through A.
goes through B,
Two
such points are
called conjugate points. (4)
pole of lines.
Also I'
if
the pole of a line
will lie
on
/,
two such
I
lie
lines
on another line
I',
the
being called conjugate
THE CONIC SECTIONS
93
The harmonic property of the pole and polar which must hold also for the conic sections since
(5)
obtains for the circle
cross-ratios are unaltered (6)
As an
by
projection.
involution range projects into a range also in
involution, pairs of conjugate points for a conic
which
lie
along
a line will form an involution range whose double points will be
the points (7)
any) in which the line cuts the curve.
(if
Similarly pairs of conjugate lines through a point will
form an involution pencil whose double lines are the tangents (if
any) from the point. 93.
Circle projected into another circle.
The curve
of projection of a circle
is
under certain con-
ditions another circle.
Prop.
If
in the curve of projection of a circle the pairs of
conjugate lines through the point
pole of the vanishing line the curve is a circle having
P
luhicJi is the
form an orthogonal its
projection of the involution, then
centre at P.
For since the polar of P is the line at infinity, the tangents any chord through P meet at infinity. But since the involution pencil formed by the pairs of conjugate
at the extremities of
lines
P
through
is
on a line through
an orthogonal one, these tangents must meet
P
perpendicular to the chord.
Hence the
tangents at the extremities of the chord are at right angles to
it.
Thus the curve has the property that the tangent it is
That
the curve
is,
Cor.
a
point within
at every
perpendicular to the radius joining the point to P.
point of
circle it
is
a circle with
P
can be projected
as centre. into
another circle
luith
any
projected into the centre.
For we have only to project the polar of the point to infinity and the involution pencil formed by the conjugate lines through it
into an orthogonal involution.
Note.
The point
be within the circle
Note
to § 87).
to
if
be projected into the centre needs to is to be a real one (see
the projection
THE CONIC SECTIONS
94
Focus and directrix as pole and
94.
Prop.
//'"
plane of
in the
there exist a point
S
polar.
of projection of a
such that the involution pencil
the conjugate lines through its
the curve
IS
an orthogonal
is
one,
circle
formed hi/ then 8 and
polar are focus and directrix for the curve.
Let
P
and Q be any two points on the curve, and
let
the
tangents at them meet in T.
Join
S
ST
cutting
PQ
in R,
and
Join SF, and draAv PAI,
in F.
let
PQ
meet the polar of
QN perpendicular
to the polar
of S.
Then ST
S
is
the polar of F, for the polar of
since that of
S
since the polar of .'.
goes through F, and
T goes through F. SF and ST are conjugate
But by hypothesis the conjugate
lines at
F
goes through
also goes
it
*S^
through
lines.
form an orthogonal
involution.
TSF
.-.
And by
.-.
.-.
a right angle.
the harmonic property of the pole and polar
{FR,PQ) = -l. SF are the bisectors of the angle PSQ,{^ SP:SQ = FP:FQ = PM QN (by similar triangles). SP PM = SQ QA\
ST
and
:
.-.
is
:
:
T
72).
THE CONIC SECTIONS Thus the
S
polar are focus
Note. infinity
from the polar of
and directrix
If the polar of
for
S
is
constant, that
the curve.
should happen to be the line at
then the curve of projection
We may
on the curve from is S and
ratio of the distance of points
to their distance
its
95
is
a circle
here remark that the circle
(§ 93).
may be
considered to
have the focus and directrix property, the focus being at the
and the directrix the
centre, is
The
line at infinity.
eccentricity
the ratio of the radius of the circle to the infinite distance of
the points on the circle from the line at infinity.
Parallel chords.
95.
we go on
Before
to establish the focus
and directrix property
of the curves of projection of a circle, which
shewing that
for all of
them there
we
shall
do by
exists at least one point S,
the pairs of conjugate lines through which form an orthogonal involution,
we
will establish
a very important general propositi(jn
about parallel chords.
Prop. In cwy conic the locus
section (or curve of projection of a circle) of the middle points of a system of parallel chords is
a straight
line,
and
the tangents
meets the curve are parallel
at the points cohere this line
to the chords.
Moreover the tangents at the extremities of each of the jicirallel ivill intersect on the line which is the locus of the middle points of the chords, and every line parallel to the chords and in the plane of the curve is conjugate with this line containing the chords
middle jwints.
Let QQ' be one
<>f
the chords of the system and
^1/ its
middle
point.
The chords may be considered
as concurrent in a point
which is the projection of a point and we have {QQ',iMR) = -l.
infinity line
;
••
that
is
m
Thus points
M
is
so too.
R at
on the vanishing
(qq',mr)^-l
on the polar of
r.
as the locus of the points is
r
m
is
a line, that of the
THE CONIC SECTIONS
96
P be a point in which this loous meets the curve, and be the projection of p, then as the tangent at p goes through r that at P must go through R that is the tangent at Let
let
P
;
P is
parallel to QQ'.
Further as the tangents at q and q' meet on the polar of Q and Q' will meet on the projection of the polar of
those at
that
is
on the
line
which
is
7\ r,
the locus of the middle points of
the chords.
Also every line through r will have its pole on the polar of r, and therefore every line through R in the plane of the conic will have its pole on the line PM, that is, every line parallel to the chords
is
conjugate with the locus of their middle points.
In other words, the polar of every point on the line which is
the locus of the middle points of a system of parallel chords
is
a line parallel to the chords. 96.
We
Focus and directrix property established. are
now
in a position to establish the focus
and directrix
property of the conic sections, defined as the projections of a circle.
We
separately,
shall
and
take
the
parabola,
ellipse
and hyperbola
in each case prove a preliminary proposition
respecting their axes of symmetry.
Prop.
A
parabola (or the projection of a circle touched by has an axis of symmetry which
the vanishing line in its plane)
meets the curve in
tivo
points one of luhich
is at infinity.
THE CONIC SECTIONS Let the vanishing line touch the
97
circle in
co.
In the plane through V, the vertex of projection, and the vanishing line draw vanishing line in
r.
Vr at right angles to Vco, meeting the Draw the other tangent ra to the circle.
Now / is the pole of aw, and therefore if pp' be any chord which produced passes through 7- and which cuts aco in n then (pp', 7ir)
= -l.
Thus, using c<^rresponding capital letters in the projection,
and remembering that ru)
rno} will project into a right angle since
subtends a right angle at V, we shall have a chord FF' at
right angles to Ail and cutting
it
at
N so
that
{FF\ NR) = -1. But
R
is
Thus
at infinity. all
.-.
FN=^^F\
the chords perpendicular to
and the curve
is
Ail
are bisected
by
therefore symmetrical about this line, which
it,
is
called the axis of the parabola.
The and
axis meets the curve in the point A, called the vertex,
in the point
As
H
which
is
at infinity.
raay projects into a right angle (for
angle at V) the tangent at
A
is
rw subtends a right
at right angles to the axis.
Finally the curve touches the line at infinity at O.
Prop. ^4 parabola (or projection of a circle touched 97. by the vanisliing line) has the focus and directrix property, and the eccentricity A. G.
of the curve
is unity. 7
THE CONIC SECTIONS Let
P be
any point on the curve of projection.
be the chord through P, perpendicular to the
axis,
Let PNP' and cutting
it in iY.
The tangents
at
P
and P'
will intersect in a point
T
on the
line of the axis (§ 95).
Then
.'.
as
Now YS
n
is
let
the pole of I^P'
at infinity,
it
polar of
be
TA = AN.
the tangent at
at right angles to
The Let
T is
as
S
PF to
will
P
meet that at meet the axis in
A
in
]'
and draw
S.
be at right angles to the axis
(§ 95).
XM cutting the axis in X, the tangent at P in Z, and
the line through
P parallel
to the axis in
M.
Join SP, SZ.
Now
as
*S'
is
the pole of
XM
An) = -i. TA = AN we have TS = XN = MP. But TY.YP=^TA.AN=^\, so that A TYS ^APYS, and TS=PS. and the Thus PS = PM, that is P is equidistant from (xs,
.-.
XA= AS, and as
>S^
polar of S.
THE CONIC SECTIONS
ST
Further, since
SPMT is Thus
a
equal and parallel to
is
rhombus and
PT bisects
ASPZ = AMPZ,
Now Z is S
S
PM and SP = PM,
the angle
SPM.
ZSP = Z ZMP = a right Z the polar of Z must go through
and Z
the pole of SP, for
(since that of
99
.
goes through Z) and through
P
since -Z'P is
a tangent at P.
SP
Hence SZ and
are conjugate lines for the curve, and
they are at right angles. >S'
at right angles to
So
also are
ST
and the
line
through
it (§ '95).
Therefore the involution pencil formed by the pairs of con-
jugate lines through polar
XM
eccentricity 98.
are focus is
SP
Prop.
:
*S'
is
an orthogonal one.
Thus
and directrix for the curve which is unity.
*S'
(§ 94),
and its and the
PM
The projection of a its plane is either a
circle
not
met by the
a closed curve having two axes of symmetry, mutually perpendicular, on which are intercepted by the curve chords of unequal length.
vanishing line in
^\
circle or
THE CONIC SECTIONS
100
Using corresponding
capital letters in the projection,
C
have that the chord FF' through
G and
intersection with the polar of C, which
its
infinity.
Thus every chord through C
is
the line at
bisected at C.
For this
is
C
chords through
are called diameters.
it
The tangents
and the
called the centre of the curve,
is
at the extremities of
any diameter are
parallel,
the tangents at the extremities of chords of the
through
c
shall
FC =CF'.
.-.
reason the point
for
we
divided harmonically at
is
meet
in the polar of
c,
which
the vanishing
is
circle line.
First suppose that the involution pencil formed by the pairs
G
of conjugate lines through
curve
a
is
is
an orthogonal one
;
then the
circle (§ 93).
Next suppose that the involution
pencil
is
not an orthogonal
one; then there must be one and only one pair of conjugate lines
G
through
mutually at right angles
Let the curve intercept on these
Draw
the chords
FQ
and
respectively cutting -them in
FR
Then
as
fl' is
lines chords
perpendicular to
N and
points at infinity on the lines
(§ 86).
and let of A A' and BB'. Jli,
the pole oi AA', and
FQ
O
A A'
and BB'.
A A'
and BB' and H' be the
passes through O',
{FQ,Nn') = -l.
FN=NQ.
.-.
FM = MR.
Similarly
Thus the curve
AGA',
is
symmetrical about each of the two lines
BGB', which are called the
We
shall
now shew
that
Let the tangents at rectangle,
But
and
GK
GK
bisects
Hence
AA' and BB'
and
B
meet
GK an'd (§ 95).
G
is
cannot be equal.
in
K, then
GAKB is
a
AB.
bisects the chord through
every chord through
jugate lines
A
axes.
G
parallel to
AB,
for
bisected at G.
the line through
G
parallel to
AB
are con-
THE CONIC SECTIONS
lOl
would be at right angles if CA = CB. And were equal the involution pencil formed by the pairs of conjugate lines through G would be an orthogonal one which is contrary to hypothesis.
But these
thus
GA
if
lines
GB
and
;
Hence AA' and BB' cannot be
We ^^'
shall
A A' to be the greater of the two. major axis and BB' the minor axis.
Then
suppose
called the
is
equal.
Prop. An ellipse (or curve of projection, other than a of a circle not met hij the vanishing line in its plane) has the focus and directrix property and the eccentricity of the curve 99.
circle,
is less
than unity.
Assuming that the projection is not a circle we have as shewn in § 98 two axes of symmetry AA', BB' of which AA' is the greater.
With
centre
B
and radius equal S and S'.
to
GA
describe a circle
cutting the major axis in
The
polars of
Let these be the tangent at
Now
;
perpendicular to
S' are
in
F and
,
cutting
A A'
in
AA'
X
(§ 95).
and X' and
F'.
since the polar of
S
through
B
S and
XF and X' F' S
but the polar of
F
goes through F, that of
F
goes through B, since
goes
FB
tangent. .-.
SB
We
is
will
the polar of F, and SF,
SB
shew that they are mutually
are conjugate lines. at right angles.
is
a
THE CONIC SECTIONS
102 ^ince
S
the pole of A''^
is
/S'X) = -1. CS.GX = CA"-.
(^^', .-.
Now Then
draw
SK parallel
BK
.
to CB to meet BF in K. = KF OS SX = CS (CX - OS) = GA"- - CS-' =^SB'-GS'=SK' FSB is a right angle. .
.'.
THE CONIC SECTIONS Let the vanishing be the pole of the
line cut the circle in
103
w and
co' ,
and
let c
line.
Let pp be any chord the line of which passes through c. divided harmonically at c and its intersection j^P' is
Then
with WW.
Using corresponding capital letters in the projection, we have that the chord FP' through C is divided harmonically G and its intersection with the polar of C which is the line
shall
at
at infinity.
.-.
FC = CP'.
Thus every chord through C bisected at G, which
is
and the chords through Let
it
in the curve of projection is
therefore called the centre of the curve,
G
are called diameters.
be observed that not every line through
G
meets the
curve, since in the plane of the circle there are lines through c
which do not meet
Of each the
it.
pair of conjugate lines through c only one will
circle, for cut
and cw' are the double
formed by these conjugate
meet
lines of the involution
lines.
Further the involution pencil formed by the conjugate lines
G cannot be an orthogonal one, since namely the projection of c&> and cw.
through lines,
Thus there through
G
will
it
has real double
be one and only one pair of conjugate lines
mutually at right angles'
(§ 86).
THE CONIC SECTIONS
104 Let
this pair
CA
be
and GB, of which the former A and A'.
is
the one
that meets the curve, namely in
Note that the curve of projection will have two tangents, G whose points of contact fl and iV the projections of co and co' are at infinity. These tangents are called asymptotes. from
CH
Since
and GDf are the double
lines of the involution
pencil formed by the conjugate lines through
G {mr, AB)=--l .•.
GA
Since
GB
and
bisectors of the angles
are
G
(§82).
at right
between Of! and
angles,
they are the
GVL' (§ 72).
To prove that the curve is symmetrical about GA and GB we draw chords PQ, PR perpendicular to them and cutting them in and M. Let Z and Z' be the points at infinity along the lines GA and GB. Then since Z' is the pole of AA' and
N
PQ
passes through Z',
(PQ, NZ') .-.
Similarly
= -1.
PN = NQ.
PM = MR. of symmetry mutually at right which meets the curve, and the other not. AA'
Thus the curve has two axes angles, one of
which meets the curve
is
called the transverse axis
and
GB
is
called the conjugate axis.
Note.
We
At present
B
is
not a definite point on the line GB.
shall find it convenient later
point to be emphasised
the curve, and
is
on to make
it
definite
;
the
that the transverse axis does not cut
we cannot determine
points
B
and B' on
it
as
these points are determined in the case of the ellipse. 101.
Prop.
A
hyperbola (or curve of projection of a circle and directiHx property,
cut by the vanishing line) has the focus
and
the eccentricity is g7-eater than unity.
Using the notation of the preceding article, we describe a circle with centre G, and radius GA cutting GH in and L', and Gfl' in K' and L, as in the figure.
K
The
lines
KL
and K'L' will be perpendicular to the transwe have seen, GA bisects the angle OCH'.
verse axis, since, as
THE COXIC SECTIONS the poles of these
.•.
on the
will lie
We
will
KL
which we
now shew
S and and K'L'.
that
and K'L' cut
AA'
in
its
{AA',
.'.
KCl
is
.•.
S and S\
polar
KL
are focus
and
of the pole and polar
SX) = -
1.
CS.CX = CA'=CK\ CKS is a right angle.
.-.
Now
denote by
X and A".
Then by the harmonic property
K
must go through S, since that of S goes Moreover the polar of goes through H, since
the polar of
through K.
will
line of the transverse axis (§ 95).
directrix, as are also S'
Let
lines,
105
K
a tangent at O.
Sn
is
the polar of
A",
that
is >S7i
and SD. are conjugate
lines.
But Sfl
is
n
being at infinity SCI
is
parallel
to KQ., that is
perpendicular to SK.
Thus we have two
pairs of conjugate
lines
mutually at right angles, namely Sfl and SK, and through S at right angles to
line
Hence the
SC
through >S'6'
S,
and the
(§ 95).
by the pairs of conjugate lines and its polar through S is an orthogonal one, and therefore are focus and directrix for the curve. pencil formed
*S'
KX
THE CONIC SECTIONS
106
Similarly S' and K'L' are focus and directrix.
The
eccentricity
Sn
:
the ratio
is
perpendicular from
II
on
KL
= KVl the same = GK OX = CA CX which is greater than Note too that as C^. CX = CA'^, the eccentricity also = :
:
:
unity.
C'>S^:
We
might have obtained the eccentricity thus
It is the ratio
C'J..
:
SA AX = CS- CA CA - CX = CH CX - CA CX CX {CA - CX) = CA (CA - CX) CX (CA - CX) = CA CX. :
:
.
.
:
:
:
102.
Central and non-central conies.
We
have seen that the
We
have proved in
Diameters.
and hyperbola have each a centre, that is a point such that ever}^ chord passing through the same is bisected by it. Ellipses and hyperbolas then are classified together as central conies. The parabola has no centre and is called non-central. §
ellipse
95 that the locus of the middle points
of a system of parallel chords
is
a straight
case of the central conies this line for the
Clearly in the
line.
must go through the
diameter parallel to the chords
is
centre,
bisected at that point.
In the case of the parabola, the line which
is
the locus of the
middle points of a system of parallel chords is parallel to the axis. For such a system is the projection of chords of the circle concurrent at a point r on the vanishing line; and the polar of
r,
which projects into the locus of the middle points of the system of chords, passes through ro the point of contact of the circle with Thus the locus of the middle points of the the vanishing line. system of parallel chords of the parabola passes through fl, that is,
the line
is
parallel to the axis.
All lines then in the plane of a parabola
and
parallel to its
axis will bisect each a system of parallel chords.
are conveniently called diameters of the parabola.
These
lines
They are
not diameters in the same sense in which the diameters of a
THE CONIC SECTIONS
107
central conic are, for they are not limited in length
and bisected
at a definite" point.
Ordinates of diameters.
103.
Def
The
.
parallel chords of a conic bisected
by a particular
diameter are called double ordinates of that diameter, and the half chord
The
is
called
an ordinate of the diameter.
ordinates of a diameter are as
we have seen
parallel to
the tangents at the point or points in which the diameter meets the curve.
The
ordinates of an axis of a conic are perpendicular to
that axis.
The ordinates
of the axis of a parabola, of the major axis of and of the transverse axis of a hyperbola are often called simply ordinates without specifying that whereto they are ordinates. Thus the ordinate of a point P on a parabola, ellipse or hyperbola must be understood to mean the perpendicular on the axis, the major axis, or the transverse
an
ellipse,
'
'
PN
axis, as
the case
may
be.
Note. When we speak of the axis of a conic there can be no ambiguity in the case of a parabola, but in the case of the ellipse and hyperbola, which have two axes of symmetry, there would be ambiguity unless we determined beforehand which axis was meant. Let it then be understood that by the axis of
a conic will be 104.
meant that one on which the
The contents
importance
for
a
right
foci lie.
of the present chapter are of great
understanding of the conic sections.
The student should now have a good general idea of the form of the curves, and, as it were, see them whole, realising that they have been obtained by projecting a circle from one plane on to another.
which
all
We
shall in the
next chapter set forth properties
common, and in subsequent chapters ellipse and hyperbola separately, shewing
conies have in
treat of the parabola,
the special properties which each curve has.
108
CHAPTER X PROPERTIES COMMON TO ALL CONICS Proposition.
105.
P
joining two points
S
If the line {produced if necessary) and Q of a, conic meet a directrix in F, and
be the corresponding focus,
between
SP
Fig.
if e
will bisect one of the angles
1.
For, drawing
have,
SF
and SQ.
PM and
QR
perpendicular to the directrix
be the eccentricity,
SP:PM=e = SQ:QR, SP:SQ = PM:QR .-.
= FP:FQ
(by similar
As FQR, FPM).
we
PROPERTIES COMMON TO ALL CONICS .-.
in
fig.
in figs 1 2
and
3, *S'^
bisects the exterior angle of
bisects the angle
it
109
PSQ
PSQ, and
itself.
We see then that SF bisects the exterior angle of PSQ, if and Q be on the same branch of the curve, and the angle P»SQ itself if P and Q be on opposite branches.
P
Prop. //" the tangent to a conic at a point P meet a Z and S he the corresiDonding focus, ZSP is a ricfht
106.
directrix in angle.
easily seen from the following considerations
This
is
The
focus and directrix are
'
pole and polar
'
for the conic,
therefore the tangents at the extremities of a focal chord will
meet
at
Z in
Thus SZ and lies
on SZ.
the directrix, and
SP
conjugate
Z
will
PSQ
be the pole of PQ.
lines, since
the pole of
SP
PROPERTIES COMMON TO ALL CONICS
110
But the
But plane
pairs of conjugate lines through a focus are at right
ZSP
Therefore
angles.
we
as
curve
projection of
is
a right angle.
are going to prove in the next article that every
having the focus and directrix property
some
circle,
we
the
is
another proof of the
will give
proposition dependent only on this property.
Regard the tangent the chord
PP' when P'
at
P
is
very close to P.
as the limiting case of the line of
angle of
PP' meet the directrix in F, SF bisects the exterior PSP' (§ 105) for P and P' are on the same branch.
And
the nearer P' approaches P, the more does this exterior
Now
if
angle approximate to two right angles.
Thus Z
ZSP =
the limit of
=a
FSP' when
P' approaches
P
right angle.
should be observed
It
that
this second proof yields also the
result that tangents at the ex-
tremities of a focal chord intersect in the directrix, since the tangent at either is
end of the chord
determined by drawing
right angles to the chord lo
PSQ SZ at meet
the directrix in Z; then ZP,
ZQ
are the tanoents.
In the preceding chapter we defined the conic sections
107.
as the curves of projection of a circle
and showed that they
We
shall
now
establish
2)lane curve having the focus
and
directrix
have the focus and directrix property. the converse proposition.
Prop. property
Every
is the projection
of some
circle.
For we have shown in the second
jjart of §
having the focus and directrix property
is
106 that a curve
such that tangents at
the extremities of any chord through S, the focus, intersect on the directrix on a line through
8
perpendicular to the chord.
PROPERTIES COMMON TO ALL CONICS
Now
project so that the directrix
is
so that the orthogonal involution at
orthogonal involution
Then the curve
111
the vanishing line
S
(§ 87).
of projection has the property that the
tangents at the extremities of every chord through jection of S,
meet
and
;
projects into another
s,
the pro-
at infinity on a line through s perpendicular
to the chord.
Hence the tangent
at each point of the curve is at right
angles to the radius joining the point to
curve
is
a
It follows of course
108.
s,
and therefore the
circle Avith s as centre.
from what we have established
the curve having the focus and directrix property had its eccentricity unity then the circle into which it has been projected must touch the vanishing line in the preceding chapter that if
in the plane
then the tricity
circle
the circle
;
if
the eccentricity be less than unity
does not meet the vanishing line
be greater than unity the circle
;
if
the eccen-
cut by the vanishing '
line.
109.
is
Pair of tangents.
J/ a pair of ttDiyents TF, TQ be drawn to a conic P and S be a focus, then SP and i^Q make equal angles with ST, and if PQ meet the corresponding directrix in F
Prop.
from a point Z
TSF is
a right angle.
Let Tti meet
FQ
in R.
TROPERTIES COMMON TO ALL CONICS
112
PQ is the polar of T, and F must go through T. since F is on the directrix
Since polar of
But through
this goes
through F, the
the polar of
F
must go
8.
Thus 8T
is
the polar of F.
8F
and 8T are conjugate lines, and as they are through a focus, they must be at right angles.
Hence
Further
Thus
SP
and
SR
SQ
Note.
and
SF
are the bisectors of the angles
It will
be seen that
T lie
PSQ, but
if
the points of contact of the
on the same branch of the curve if
ST
bisects
they are on different branches then
bisects the exterior angle of
The
between
(§ 72).
tangents from the angle
FR) = -l S(PQ,FR) = -1.
(PQ, .-.
figures given do not
ST
PSQ. shew the case v/here
both touch the branch remote from
S.
TP
and
The student can
TQ
easily
represent this in a figure of his own. 110.
The above
proposition gives a simple construction for
drawing two tangents to a conic from an external point
T.
PROPERTIES COMMON TO ALL CONICS
ST
Join
and
meet the conic
let it
in
K and K'. Take
of the preceding article can be utilised.
KK') = -
that {TR,
TP
TQ
and
.-.
the pole of
But the pole .
.
.
.
are at right angles
such
the directrix in F.
P and
Q.
and through a
focus,
on SF.
/SfT lies
of SI'
on the
is
T goes
the polar of
But the polar Thus the line FR
figure
KK'
lines.
the pole of ST, that
i^ is
The in
are the tangents.
ST and SF
For as
they are conjugate
R
1 (§ 70, Cor. 2).
Draw SF at right angles to ST to meet Draw the line FR and let it cut the conic in Then
113
of
T
directrix.
the polar of
is,
goes through
R
since (TR,
the polar of T, that
is
F goes
through T.
through F.
is,
PQ
is
KK') = -
1.
the chord of
contact of the tangents from T.
The Normal.
111.
Def.
The
line
and at right angles
Prop. and
S
be
If
the
through the point of contact of a tangent to it is called the
normal
a focus of the
P
normal at that
point.
any conic meet the aocis in G, conic then SG = e SP, where e is the at
to
.
eccentricity.
Let the tangent at
P
meet the
directrix corresponding to
SinZ.
Draw
PM perpendicular to the directrix.
Then
since
A. G.
PM is'Jparallel to the axis, Z MPS = Z PSG. 8
PROPERTIES COMMON TO ALL CONICS
114
PMZ and i^S'Z are right angles, PSZM is SMP = Z SZP = complement of z SPZ = z SPO. Thus the A s ISPG, PMS are similar and SG:SP = PS:PM = e- .-.80 = 6. SP. Also since
and
cyclic
^^
The student can make
P
where
this case
is
it will
So that the 112.
The The
In
latus rectum. focal
chord perpendicular to the axis on which
lies is called
Thus the
latus
focus to the axis
(§
the latus rectum of the conic.
rectum
the double ordinate through the
is
103),
The semi-latus rectum of a conic
Prop.
S.
be found
ZSMP = 180°- ZSZP = 90°+ ZSPZ= ZSPG. and PMS are still similar. As SPG
Def. the focus
himself a figure shewing the case
for
on the branch of a hyperbola remote from
is
aharmonic mean
between the segments of any focal chord.
Let
aS7>
Draw and
be the semi-latus rectum, and
PM and QR perpendicular
QK perpendicular to Then
.
PSQ
any
focal chord.
to the directrix,
and
PN
the axis.
SP:P3f = e = SL:SX = 8Q QR. :
And by
similar triangles
SP .SQ = SN: KS = XN - XS XS - XK (Fig. = MP - XS A^S' - RQ = e{MP-XS):e(XS-RQ) = SP- SL SL - SQ. :
1
:
:
..
SP,
SL and SQ
are in harmonic progression
and
A
_L J_ SP'^SQ'~SL' This proposition requires some modification on opposite branches. We ijow have
if
P
and Q are
PROPERTIES COMMON TO ALL CONICS
115
XR :KX + XS (Fig. = e{XS-3IP):e(QR + XS) = SL-SP:SQ + SL
SP SQ = SN KS = XS :
:
Fig. .-.
2)
Fig. 2.
1.
SP {SQ + SL) = SQ (SL - SP) SQ SL - SP SL = 2.ST SQ .
.
.
.
.
1 __L_ ^
''
Thus
in this case
Cor.
it is
IVie rectangle
SP,
SP
SQ'SL'
SL and — SQ
contained by
tJie
that are
m
h.p.
segments of any focal
chord varies as the length of the chord. ^
- SQ
SP
SL
'
P
and Q be on the same or opposite branches {P being on the branch adjacent to S), and in both cases we PQ ^ _2_ have
according as
SP SQ .
.-.
that
SL
SP.SQ = '^^'xPQ SP.SQozPQ.
is
and SQ are in opposite directions P and Q If same branch of the curve, and if they are in the same P and Q lie on opposite branches.
SP
Prop. Any conic can be projected 113. any point in the plane of the conic projected
lie
on the
direction,
into a circle loith into the centre
the circle.
8—2
of
PROPERTIES COMMON TO ALL CONICS
116 For
let
P
Take the
be any point in the plane of the conic. jiolar of
P
vanishing line and project so>
for the
that the involution pencil formed by the pairs of conjugate
P
through
lines
projects into an orthogonal involution
then exactly as in is
a
§
(§
87);
98 we can prove that the curve of projection
circle.
Note. that
is
P
may be
In order that the projection
P
tangents from
must
to the conic
lie
must not be
a real one, the
(Note to
real
§ 87),
within the conic.
Carnot's theorem.
Prop.
114.
If a
conic cut the sides of a triangle
ABC in
A^, A^; B,, B^; C\, C^; then
AB, AB^ CA, CA, BC, BC, = AC\.Aa. BA, BA, CB, .
.
.
.
.
.
.
Project the conic into a circle
by corresponding small
projection
Then
since
abi
.
C«i
.
bcy .' .
.•.
ahi
.
ah,
.
ca-i
.
ca.,
•
bci
ah =
.
.
letters.
ac^
= c6i
bc2
=
.
.
buy.
= aCj
.
ac2, cb.2 ba.2,,
ac.^
the triangle formed by the lines
spective with the triangle abc
CB,.
and denote the points in the
;
COa
bc.2
.
(§ 68).
.
bui
.
ba^
.
ttj^a, b^Co,
cb^
.
063.
c^a^ is in
per-
PROPERTIES COMMON TO ALL CONICS .•.
117
the triangle formed by the lines A^Bo, B^Co, G^Ao
is
in
ABC.
perspective with the triangle
.-.by §68
AB, AB, CA, .
.
.
CA.^
.
BC, BC, .
= AC\.ACo_. BA, BA, .
.
CB,
.
CB.,.
Newton's theorem. 115.
and PQ,
Prop.
RS be
If
a variable
be
jjoint in the
plane of a
conic,
chords in fixed directions through 0, then
OP.OQ OB. OS
IS constant.
Let 0' be [any other point and through 0' draw the chords PQ and RS.
P'Q', R'S' parallel respectively to
Let QP, Q'P' meet in Let P'Q' and
Now
RS
at infinity
and SR, S'R'
in
CI.
in T.
apply Carnot's theorem to the triangle coOT and get ayP oyQ .
coP'
but
&>
meet
.
(oP
OR OS TP' TQ' TR TS OP OQ .
.
o)Q'
.
~p;=
loP
.
,
1
.
.
.
,
&>(
.
and ~^. -1. o)Q
1,
PROPERTIES COMMON TO ALL CONICS
118
TR. TS ~ OK.OIS' Next apply Carnot's theorem
to the triangle D.TO'
and get
n R.nS. TF. TQ O'R O'S' ~ nw ns' o'P' O'Q' tr Ts '
.
.
.
.
.
.
TP'.TQ' ^ O'F'
TR. TS ~ O'R' O'S' OP OQ _0'P' .O'Q
•
•
.O'Q' •
.
.
Hence ,L
.
that
IS,
OR.OS~
OP.OQ 77^5—7T-< UK (Jo
O'R'.O'S"
.
IS
constant.
.
This proposition
is
known
as
Newton's theorem.
Note. In applying Newton's theorem it must be remembered that the lines OP, OQ, &c. have sign as well as magnitude. If OP and OQ are opposite in direction, they have opposite sign, and
OR
so for
and OS.
Newton's theorem
116.
see in later chapters,
We
is
of great importance, as
where considerable use
give some propositions illustrating
will
If two chords of a conic PP' and QQ', OP' OQ OQ' is equal to that of the focal chords parallel to PP' and QQ'.
OP
Let the
.
:
shall it.
its use.
Prop.
the ratio
we
be made of
.
intersect in
lengths of the
PP' and QQ' be pSp' and
focal chords parallel to
qSq'.
Then by Newton's theorem
OP OF OQ .
:
.
OQ'
In the special case where
have OP'
= - OP
and OQ' .-.
We
Note.
=-
= Sp Sp Sq Sq' = pp':qq (§ 112 Cor.). .
is
:
.
the centre of the conic
we
OQ.
OP"-:OQ^-=pp':qq'.
have already explained that in using Newton's
theorem, the signs of the segments of the line are to be considered. If *S^
.
OP
.
Sj^ and
OP' and OQ OQ' have opposite signs so also will Sq Sq' have opposite signs. This only happens in .
.
PROPERTIES COMMON TO ALL CONICS the case of the hyperbola, and
when one
119
of the four points p, p'
Then q lies on the opposite branch to the other three. one of the focal chords pp', qq will join two points on opposite and
q,
branches and the other will join two points on the same branch. If
we make the convention that a negative value be attached it joins two points on opposite
to the length of a focal chord if
branches otherwise
it is
to count positive the relation
OP.OP:OQ.OQ'^pp:qq' is
algebraically as well as numerically correct.
So
also
it is
true, with the
same convention
as to sign, that if
CQ
be the semidiameters parallel to the focal chords, pp', qq'
And
from this we see that of the two diameters parallel to
CP,
pp' qq' :
two
focal chords,
= CP' CQ\ :
one of which joins two points on the same
branch and the other two points on opposite branches, only one can meet the curve in real points for the ratio CP-
:
CQ- has now
a negative value.
Prop. // OP and OQ be two tangents to a conic OP' OQ- is equal to the ratio of the focal chords parallel respectiveli/ to OP and OQ. 117.
then
:
Let the
focal
chords be pSp', qSq'.
Then regarding
OQ
meeting the curve in two coincident points P, and we have by Newton's theorem
OP.OP= OQ .OQ = Sp. Sp
Sq
.
Sq',
OP\:OQ^=pp:qq'.
.-.
Whence we
:
OP as
similarly,
see that the focal chords have the
same
sign.
from a point to equal to that of the diameters parallel to them.
It is clear too that the ratio of the tangents
a central conic 118.
is
Prop.
//"
a
circle
cut
a conic in four points the
chords joining their points of intersection in pairs are equally inclined to the axis.
Let the conic and
circle intersect in the four points
P,
Q, P', Q'.
Let PP' and QQ' intersect in 0.
Draw
focal chords pSp',
qSq
parallel to
PP' and
QQ'.
PROPERTIES COMMON TO ALL CONICS
120
Then, by Newton's theorem
OP.OP'-.OQ. OQ' = Sp
.
Sp'
:
Sq
.
Sq'
= pp':qq'. But from the
circle
op.OF = OQ.oq, ^^d ^P ^P' = ^9 ^'j'chords have the same sign, their lengths are ecjual and the rectangles contained by their segments
PP =
•
'
Thus the
•
parallel
'Vl'
•
•
focal
are equal.
These
choi'ds
must then be symmetrically placed and make Thus PP' and QQ', parallel to them,
equal angles with the axis.
make
equal angles with the axis.
CoR.
two
If a circle touch a conic at one point
others, the tangent at the point of contact
joining the two points of intersection
make
and cut it at and the chord
equal angles with
the axis.
Circle of curvature. 119.
An
infinite
number
of circles can be
drawn
to touch a
conic at a given point P, such circles having their centres along
the normal at the point.
These
circles will in general cut
the
conic in two other points, but in the special case where one of
these other two points coincides with the point of contact circle is called the circle of
curvature at P.
This
circle
P
the
may be
regarded as the limiting case of the circle passing through
P
and through two points on the conic consecutive to P, so that the conic and the circle have two consecutive tangents in common. They have then the same rate of curvature at that point.
The
subject
of curvature
Differential Calculus, but
principal
properties
of
it
the
properly
belongs
to
the
seems desirable to give here the of
circles
curvature
of
conies*
Accordingly we shall at the end of the chapters on the parabola, ellipse, circles of if
and hyperbola add a proposition relating to the It is clear from § 118 that
curvature for these curves.
the circle of curvature at a point
PQ
P
of a conic cut the conic
and the tangent at P are equally inclined For the tangent at P and the chord PQ are the to the axis. common chords of the circle and the conic. again in
Q then
PROPERTIES COMMON TO ALL COXICS
The
121
following figure illustrates the circle of curvature at a
poipt of a conic.
Self-Polar Triangle. 119a.
Prop.
If a
conic pass through the
quadrangle, the diagonal or harmonic triangle
regard
to the conic
— that
is,
each vertex
is
four is
i^oints of
a
self-polar with
the pole of the opposite
Mde.
Let
ABCD be the quadrangle; PQR the diagonal or harmonic
triangle.
Let
PQ
cut
AD and BC in X
and
F.
PROPERTIES COMMON TO ALL CONICS
122
(AD,XR)=-1,
Then .'.
R
the polar of
goes thi'ough
A'"
(§
92
(5)),
{BC\YR) = -1,
and
the polar of
.'.
.'.
Similarly
QR
is
PQ
is
goes through Y.
PR
the polar of P, and
Thus the proposition
is
of Q.
proved.
Another way of stating the diagonal points
J?
the polar of R.
this proposition
when taken
would be
to say that
in pairs are conjugate for the
conic.
The
triangle
PQR
also called self-conjugate with regard
is
to the conic.
EXERCISES Given n conic and a focus and corresponding directrix 1. shew how to draw the tangent at any point.
Given two points on a conic and a
2.
locus of the corresponding focus
is
a
directrix,
PQ
shew that the
are two chords of a conic intersecting and P'Q' meet on the polar of 0.
[Project the conic into a circle and If the tangent at the
4.
end
tangent at the nearer vertex A in If the tangent at
5.
:
of a latus
rectum LSL' meet the
T then TA =AS.
any point
of a conic
/*
meet a directrix
D
in
then
the eccentricity.
If the normal at
6.
in 0,
into the centre.]
F, and the latus rectum through the corresponding focus in
SD SP =
it,
circle.
POP' and QOQ'
3.
prove that
of
P
meet the axis in G, and GL SP, tlien PL = the semi-latus
to a conic
be perpendicular to the focal
i-adius
rectum.
if
7.
If
PSP'
PQ
and
PQ
F',
FSF'
F and
be a focal chord and Q any point on the conic and meet the directrix corresponding to the focus S in is
a right angle.
PROPERTIES COMMON TO ALL CONICS If a conic touch the sides opposite to A, £,
8.
ABO
in I), U,
F
respectively then
AD, BE,
CF
123
C
oi
a.
triangle
are concurrent.
[Use §114.]
By means
9.
is
Newton's theorem prove that
of
ordinate of a point
if
PX
P on a parabola whose vertex is A, then P on
independent of the position of
be the
PN'^
:
AJ^
the curve.
Conies are drawn through two fixed points I> and E, and
10.
are such that
shew that the
DE
subtends a constant angle at a focus of them passes through
line joining this focus to the pole of
DE
a fixed point.
The polar
11.
directrix on
of
any point with respect which bisects the
the diameter
to a conic meets a focal
chord
drawn
through the point and the corresponding focus. Pro\'e that the line joining a focus of a conic to that point correspcmding directrix at which a diameter bisecting a
12.
in
the
system of parallel chords meets
it
is
[Use
§§
95 and 106.]
13.
P
and Q are two points on
perpendicular to the chords.
a
conic,
and the diameters P and Q
bisecting the chords parallel respectively to the tangents at
meet a directrix
iu
M
and
N
;
shew that
MX
subtends at the
corresponding focus an angle equal to that between the tangents at
P and 14.
conic,
Q.
Given a focus and the corresponding directrix of a variable shew that the polar of a given point passes through a fixed
point.
Given a focus and two points of a variable conic, prove that must pass through one or other of two
15.
the corresponding directrix fixed points. If
16.
two conies
focus, a chord common to the through the point of intersection of the corre-
two conies have a common will pass
sponding directrices. If
17.
P
be any point on the tangent at a point
P
of a conic of
I'M be the perpendicular to SP, and Ty the perpendicular on the directi'ix corresponding to 6', then (Adams' theorem.) S2I TJV - e.
which
is
>S
a focus, and
if
:
18.
Given a focus of a conic and a chord through that
focus,
prove that the locus of the extremities of the coi'responding latus
rectum
is
a
circle.
PROPERTIES COMMON TO ALL CONICS
124
TP
If
19.
TQ
and
be two tangents to a conic prove that the
portion of a tangent parallel to is
PQ
TP
intercepted between
TQ
and
bisected at the point of contact.
A diameter of a
conic meets the curve in P and bisects the normal at Q, shew that the diameter through Q bisects the chord through P which is a normal at P. 20.
QR
chord
which
PQ is a PQ
21.
the pole of
and S
is
22.
P
;
is
a
the corresponding focus
AA\ BB\
If
T
is
meets a directrix in
Z
chord of a conic cutting the axis in K, and the diameter bisecting
any point on the
CC
;
PQ
prove that
TS
through
AB, 23.
AC
and
U, F,
G
PC
PC"
lie
on a straight
0.
and the conic into a
A, B, C,
BD
ZK.
be chords of a conic concurrent at 0, and
[Project to infinity the line joining of
parallel to
conic, then the points of intersection of the
straight lines BC, PA', of CA, PB', and of AB, line
is
in
F
D ;
to the point of intersection
circle.]
are four points on a conic
and the tangents at A and
AB,
;
D
in
CD
meet in E;
G
prove that
;
are col linear.
AD
[Project
and
BG
into parallel lines
and the conic into a
circle.]
24.
two
If a conic be inscribed in a quadrilateral, the line joining
of the points of contact will pass through one of the angular
points of the triangle formed by the diagonals of the quadrilateral. 25.
Prove Pascal's theorem, that
if
a hexagon be inscribed in a
conic the pairs of opposite sides meet in three collinear points. [Project the conic into a circle so that the line joining the points of intersection of
two pairs
of opposite sides is projected to infinity.]
P any point on the polar of A. The tangents from P to the conic meet a given line in Q and R. Shew that AR, PQ, and AQ, PR intersect on a 26.
^
is
a fixed point in the plane of a conic, and
fixed line.
[Project the conic into a circle having the j^rojection of
A
for
centime.]
27.
A system
fixed point,
PQ
meets 28.
of conies touch
and BD,
BC
CD
AB
and .^C at
^
and
of the conies in P, Q.
C.
D
Shew
is
a
that
in a fixed point.
If a conic pass
intersection of
meet one
AG
through the points A, B,
and BD,
C, and of the tangents at
of xlZ/and
A and
D
CD,
C,
D, the points of B and
of the tangents at
are collinear.
PROPERTIES COMMON TO ALL CONICS
125
A on a conic two and *S" are two fixed points and P a variable PS, PS' meet AI, AT in Q, Q' respectively, point on the conic shew that QQ' passes through a fixed point. Til rough
29.
AI,
AT
a fixed point
are drawn,
fixed straight lines
»S'
;
30. A-^A.2,
If a conic cut the sides
AA.2, BB.2, 31.
two
BC, CA,
AB
of a triangle
ABC
in
B^B^, C1C2, and AA^, BB^, CC^ are concurrent, then will
GC^ be concurrent.
When
a triangle
is
self-conjugate for a conic,
two and
onl}^
of its sides cut the curve in real points. 32.
Prove that of two conjugate diameters of a hyperbola, one
and only one can cut the curve in real points. [Two conjugate diameters and the line at
infinity
form a
self-
conjugate triangle.]
Given four points A, B, C, shew that in general four drawn through A, B, C having ^S" as focus and that three of the conies are hyperbolas with A, B, C not on the same branch, while the remaining conic may be an ellipse, a parabola, or a hyperbola having A, B, C on the same branch. 33.
.S',
conies can be
;
Prove that a circle can be projected into a pai\abola with 34. any given point within the circle projected into the focus. Prove that a circle can be projected into an ellipse with 35. two given points within the circle projected into the centre and a focus of the ellipse.
Prove that a
circle can be projected into a hyperbola with within the circle and another given point Q without projected respectively into a focus and the centre of the hyperbola. 36.
a given point it
P
126
CHAPTER XI THE PARABOLA The form
120.
and
in §§ 96
properties of the vertex,
S
Q
for
we have
Prop.
curve.
the focus,
for
the axis, and point, as
of the parabola has ah-eady been indicated
In this chapter we shall develop the special
97.
Throughout
A
will
stand for the
X for the intersection of the directrix with
the point at infinity along the axis, at which
seen, the parabola touches the line at infinity.
The latus rectum
=
4
Let LSL' be the latus rectum. Draw LJ\l perpendicular to Then LL' = 2LS = 2LM = 'ISX = 4>AS.
the directrix. 121.
Prop.
If
PN
be the ordinate of the point P, then
PN' = ^AS.AN. Let
PN meet the
parabola again in P', and
let
LSL' be the
latus rectum.
Then by Newton's theorem
NP
.
(§
115),
NA Na SA Sn ^NA:SA
NP' SL SL = :
.
'
.
since
:
H
is
.
at infinity.
THE PARABOLA
127
PN'-AAS' = AN':AS; PN"- = 4.AS.AA'. .-.
This
prop(jsiti(ni will later
case of a
The preceding
122.
on be seen to be only a special
more general theorem. proposition shews that a parabola
the locus of a point in a plane such that the square of
distance from a line line
The
I'.
line
I
I
is
varies as its distance from a perpendicular
the axis,
the constant of variation
To determine of the line
//
is
its
is
the parabola
the point
lies.
I'
the tangent at the vertex, and
the length of the latus rectum.
we ought If
it
may
know on which
to lie
side
on either side then
is tivo parabolas, each of which is got from the other by rotating the figure about the tangent at the vertex through two right angles.
the locus
Tangent and Normal.
123.
Prop. and
G
If the tangent
respectively,
and
(2)
The
We
first
P
meet the
axis in
T
P,
NG = 2AS.
of these properties has been already proved in § 97,
have seen that
tangents at
and normal at
PN be the ordinate of TA=AN (1)
P and
if
PJ^ meet the curve again
P' meet on the line of the
in P', the
axis, that
is,
they
THE PARABOLA
128
Then by the harmonic property
intersect in T.
{TN,
polar .-.
Also since
TPG
is
PN'-
.-.
Def.
NG is
a right angle
=
TN NG = ^AN .
124.
Prop.
AN
.
NG.
{ll^l),
NG = 2AS. called the suhnormal of the point P.
a parabola the subnormal
equal angles
and
TA=AN.
PN'-=4.AS.
But
of the pole
Aa) = -l;
is
Thus
in
constant.
The tangent at any point of a 'parabola makes axis and the focal distance of the
luith the
M
Z
129
THE PARABOLA
Now
since
MPZ which
SP = PM,
and
have the angles at
PZ
is
common
to the
As SPZ,
M and 8 right angles,
A SPZ = A MPZ SPT = Z TPM =zSTP.
.-.
and Z
P meet
If the normal at
Cor.
the axis
G
then
SG = SP = ST. That ,ST = and
SP
follows from the equality of the angles
SPT
STP. Further the complements of these angles must be equal .-.
zSPG = ^8GP; .-.
We
SG =
SP.
SG
note that the equality of
follows from the fact that for
and
SP
in a parabola
SG = e SP
any conic
.
(§ 111).
Prop. The foot of the perpendicular F from the 125. focus on to the tangent at any point P of a parabola lies on the tangent at the vertex
The
We
first
and SY-
can also prove
it
thus
Let the tangent at /SF
= SA
.
part of this proposition
will bisect
P
SP. is
implicitly proved in § 97.
:
meet the
axis in T.
-Since
ST=SP,
TP.
But if PX be the ordinate of P, TA = AN. .'. ^P^is parallel to XP, that is J. F is the tangent
at the
vertex. A. G.
9
THE PARABOLA
130 Further as SY2^
ST, SY'
is
a right angle and
YA
perpendicular to
= SA ST = SA SF. .
.
Cor.
1.
Cor.
2.
zSPY=zSYA. If the locus of the foot of the perpendicular from
a fixed point on a variable line be a straight variable line touches a parabola having
line,
then the
focus at the fixed
its
point.
Def.
When
a line moves in a plane so as always to touch
a certain curve, the curve
is
called the envelope of the line.
Pair of Tangents. Prop.
126.
Tangents
to
a parabola at
tJte
extremities of a
focal chord intersect at right angles in the directrix.
That they intersect in the
directrix
we know
already, since
the focus and directrix are pole and polar.
Let
PSQ
be a focal chord, and
directrix in Z.
Then
as
let
the tangents meet in the
Draw Pif and QF perpendicular
we have seen
in §
to the directrix.
124
ASFZ^AMFZ, .-.
Si milarly
Thus
z.SZF = aMZF. Z
Z
SZQ = Z FZQ.
QZF = i =a
of
two right Z
right angle.
s
THE PARABOLA
We
If TP and TQ be two SPT, STQ are similar.
Prop.
127.
the triangles
know
already that the angles
131 tangents to a parabola,
PST, TSQ
are
equal
P and Q meet the tangent at the S7T and SZT are right angles (§ 125). Z>^PV=Z8YA (§125) = Z STZ since SZYT is cyclic.
vertex
(§ 109).
Let the tangents at in
Y and Then
Z, then
ZSPT = ZSTQ;
Thus .-.
and
the remaining angle
STQ Cor.
Z.
>S'Qrand the triangles tiPT
ST' = SP.SQ SP ST = ST SQ. TP' TQ' = SP SQ
1.
fur
Cor.
STP =
are similar.
2.
:
:
:
for
TP'
:
:
7Y,)'^
i
= ASPT A,STQ = SP.ST:ST.SQ :
since
ZPST = Z.TSQ
= SP:SQ. 128.
parabola
Prop. is
equal
TAe exterior angle between to
ttvo
tangents to a
half the angle luhich their chord of contact
subtends at the focus.
THE PARABOLA
132
the axis in
P and Q meet F and K respectively.
Then
FTK = Z SKQ - Z 8FP
Let the tangents at
Z
in T,
and
let
them meet
= zSQK-zSPT
= 2 right angles - Z SQT - Z SPT = 2 right angles - Z STP - Z ^PT = ZT.SP = 1 zPSQ. Parabola escribed to a triangle.
129.
When triangle
the sides of a triangle are tangents to a parabola, the
is
said to circumscribe the parabola.
But
it
must be
clearly understood that the triangle does not enclose the parabola, for
no
finite triangle
extent. is really
When
can enclose a parabola, which
is
infinite in
a triangle circumscribes a parabola, the parabola
escribed to
and the other two
it,
that
is, it
touches one side of the triangle
sides produced.
Only
triangles
which have
the line at infinity for one of their sides can enclose the parabola,
and
in the strict sense of the
word be said
to circumscribe
it.
however to extend the meaning of the word and to understand by a triangle circumscribing
It is convenient '
circumscribe
'
a conic a triangle whose sides touch the conic whether the triangle encloses the conic or not. 130.
Prop.
three tangents to
The circumcircle of the triangle formed by a j^arabola passes througli the focus.
THE PARABOLA Understanding the word §
may
129 we
If a
'
133 '
as explained in
its
circumcircle goes
circumscribe
state this proposition thus
triangle circumscribe a parabola
through the focus.
This can be seen from the fact that the feet of the perpendiculars from the focus on the three sides of the triangle which
touch the parabola are collinear, lying as they do on the tangent at A. .'.
S
lies
on the circumcircle of the triangle
Or we may prove the
(§ 7).
proposition in another
Let the tangents at P, Q,
R
way
form the triangle
TLM as
in
the figure.
Then
And
as as
ASPL is similar to A SLR, z SLR = zSPL. ASPT is similar to ASTQ, Z STQ = z SPT.
zSLM= zSTM,
.-.
that
is,
SLTM is cyclic, or S lies on the circle through
T,
L and M.
The orthocentre of a triangle circumscribing a
Cor.
j'jora-
bola lies on the directrix.
For centre
if
is
'TLM be the
triangle, the line joining
have seen, the pedal line of .•.
S
to the ortho-
bisected by the tangent at the vertex, which
the orthocentre must
*S (§ 8).
lie
on the
directrix.
is,
as
we
THE PARABOLA
134
Prop.
131.
triangle
SLM
is
P
tangents at
//' the
meet in T, and a third tangent at
R
cut
and Q them in
SPT and PL.LT= TM MQ = LR RM. similar
to the triangles :
By TLM.
the preceding proposition
to
a parabola
L
and M, STQ, and
the
:
8
lies
on the circumcircle of
zSML= Z.STL ZSLM = ^ SPT from the similar As SPL, SLR. A SLM is similar to A SPT and therefore also to A STQ. Further A SLR is similar to ASTM for ZSLM=ZSTM, .-.
and
.-.
SRL = Z SLP = 180° - z SLT = z SMT.
Z
and
.-.
LR:TM=SR:SM = MR MQ LR RM = TM MQ. :
.-.
Similarly
Hence 132.
(by similar
As SRM, SMQ).
:
:
MR .RL = TL: LP.
PL LT = TM MQ = LR RM. :
:
:
Diameters.
We
have already explained in § 102 what is meant by diameters of a parabola. Every line parallel to the axis is a diameter, and every diameter bisects a system of parallel chords. It
must be remembered too that the tangents at the extremities
THE PARABOLA
135
of each of the parallel chords bisected by a diameter intersect
on that diameter
Prop.
(§ 95).
TQ and TQ'
//
be the diameter bisecting
then
be tangents to
QQ' in
V
and
a parabola, and
TV P
cutting the curve in
TP = PV.
PV being parallel
For
.-.
Note that 133.
TA =
Prop.
through O.
to the axis goes
Thus by the harmonic property
of the pole and polar
{TV,Pn) = -l, TP = PV.
AX (§
123)
is
only a special case of
this.
The length of any focal chord of a parabola
of the focus from the point where the diameter bisecting the chord meets the curve.
is
four times
the distance
Let RSR' be any bisects
it
focal
chord,
PV
the diameter
which
in V.
Let the tangents at R and R' meet in Z. Then Z is both on the directrix and on the line of the diameter PV. Also
ZP=PF(§ 132). But ZSV is a right
angle
.-.
Now
draw
RM and
(§
106).
SP = PV = ZP.
R'M' perpendicular
to
the directrix.
THE PARABOLA
136
Then
2
VZ=RM + R'M' since V
is
the middle point of
RR'
= RS + 8R' = RR'; .-.
RR' = 4.PV=4.SP.
A focal chord bisected by a particular diameter is called the pcwameter of the diameter which bisects the chord. Thus RR' is the parameter of the diameter PV, and we have proved it equal to ^SP. In particular the latus rectum is the parameter of the axis,
134.
and we proved
Prop.
it
QV he
//
equal to
4>S^^.
an ordinate of a diameter
PV then
QV = 4aSP P V. .
QF
Produce the ordinate .-.
meet the curve again
in Q'.
parallel to the chord QQ'.
RR'
to
QV=VQ'. Draw
will
the focal chord
The diameter
RSR'
PV in
be bisected by
tl (say).
PV meets
the curve again in
D, at
an
infinity,
thus by Newton's theorem
VQ .-.
.
VQ. VQ'
.VP.Vn=UR.UR':UP. Un UR UR'= VP. Vn.UP.Un=VP:
VQ' :
;
.
.-.
QV':RU^-=PV:PU:
UP.
THE PARABOLA
FV
PU
137
(§
SP
133)
= 4SP. .-.
QV"-
= ^SP.PV.
It will be seen that the property is
FN- = 4 AS.
AS
of § 121
only a special case of the general proposition just proved.
135. The preceding proposition shews that a parabola may be regarded as the locus of a point in a plane such that the
square of
its distance from a fixed line I varies as its distance from another fixed line I' not necessarily at right angles to /.
The
line
I
is
a diameter of the parabola and
the point where
/
and
If a be the angle
I'
is
a tangent at
makes with the
axis in § 134
which
QV
QF=perp. from*Q on and .*.
I'
intersect.
PF x
cosec
a.
PF= perp.
from Q on tangent at P x cosec Q on PV)- x cosec- a = 4
a;
(Perp. from
Q on PF)on tangent at
(Perp. from
rerp. from
where »SF
is
y
,
cd
•
i
cosec
ot-
P
the perpendicular firom
*S^
on the tangent at P.
a.
THE PARABOLA
138
Thus
if
move
a point
distance from a line
I
is
in a plane so that the square of its
k times
distance from another line
its
k being constant, the locus of the point axis parallel to
distant
-r
tion of
I
with
from
and
it,
I'
The
it.
The
I.
and
it
also lies in a line
is
I'
of intersection with
the tangent to the parabola at
is
makes point
^122
in
if
the axis itself and
I'
the lines
I
and V are at
the tangent at the vertex.
Circle of curvature.
136.
We
/
I
its
I.
As already explained right angles
and
I'
through the intersec-
the same angle that
I'
i',
a parabola having its
focus lies in a line parallel to
and making with
line
is
have explained in
§
119 what we mean by the
We
curvature at any point of a conic. circle of
shall
circle of
now shew how
the
curvature at any point of a parabola can be determined.
The centre of the circle of curvature at P lies on the nonnal we can find the length of the chord through P of
at P, if then this circle in circle
of this chord
then
any
direction, the length of the
PD
will
and perpendicular
meet the normal
Prop. The chords of and through the focus
the circle
It is clear that these
two chords must be equal
make equal
Now
P
in D,
at
of curvature parallel to the P of a parabola = 4>SP.
any point
for
they
angles with the tangent at P.
consider a circle touching the parabola at
again at a near point Q.
through at
to it to
be the diameter of the circle of curvature.
axis
it
diameter of the
can be found by drawing a line through the other extremity
Q and let Draw
in R.
Then from the
it
Draw
meet the
QV the circle
P and cutting
the diameter of the parabola
circle
again in
K and the tangent
ordinate to the diameter
we have
RQ.RK = RP"'; RP"'
07^
PV.
THE PARABOLA
Now
in the limit
cides with P,
P
through
when Q moves up
139 to
RK becomes the chord of
and ultimately coin-
the circle of curvature
parallel to the axis.
Hence
this chord is of length 4
CoR.
The diameter
of the circle of curvature at
where
-r^.
=
Diameter
cos
(
Z between normal and
= sin Z = sin Z
P=
AiSP-o-r>-,
ox
For axis)
(
between tangent and axis)
{
between tangent and SP)
_SY
~ spNote.
The diameter of the
circle of
called the diameter of curvature,
curvature through
P are
curvature
is
commonly
and the chords of the
circle of
called simply chords of curvature.
THE PARABOLA
140
EXERCISES PSP'
1.
and P'N' ordinates
a focal chord of a paralwla, Pi^'
is
to the axis, prove
PN. P'N' =
A
2.
series of parabolas
4.
AS''
touch a given line and have a
on a
focus, prove that their vertices lie
common
circle.
a straight line rotate about a point in a plane containing it at a given
If
3.
= iAN. AN'.
the line the directions of motion of each point in
moment
are tangents to a parabola.
The ordinates
4.
on a parabola are divided in a given
of points
prove that the locus of the points dividing them
ratio,
is
another
parabola.
The
5.
is also
locus of the middle points of focal chords of a parabola
a parabola, whose latus rectum
half that of the original
is
parabola. li
6.
QV
be an ordinate of
If tlie
7.
normal at
P
PV
the diameter
QD^ =
perpendicular to the diameter
4:AS
.
and
meet the axis
to a parabola
QD
be
PV. in
G
then
PG"^iAS.SP.
PR
PQ,
8.
are
R
diameter through
through Q in
E
two chords in
prove that
;
a parabola;
of
the point
EF is
F
and
PE
PQ
meets the
meets the diameter
parallel to the tangent at P.
[Project the parabola into a circle with
E
projected into the
centre.] 9.
If a circle touch a parabola at
diameters through
Q and
A' will
P
and cut
meet the
circle
it
at
Q and
R, the
again in points on a
line parallel to the tangent at P.
[Use§ 116 10.
variable
PL
:
TM
11.
If
Cor.]
TP
and
TQ
be two fixed tangents to a parabola, and a
M
respectively the tangent cut them in L and are equal and^constant. and TL :
If
two tangents
tangent in L and
ratios
QM
TP
and
TQ
to a parabola be cutl)y a third
M respectively then TL
TP
"^
TM_~ TO
THE PARABOLA If the
12.
P
normal at
PN be the
Q and
to a parabola
141 meet the curve again in of PQ^
T the pole Pq:PT = PN:AN.
ordinate of P, and
13. If the tangent at P meet the directrix in rectum produced in D, then SD ~ SZ. 14.
through
Z and
the latus
TP and TQ are tangents to a parabola, and the diameters Pand Q meet any line drawn through Tin J/ and N prove TM- = TX- = TP TQ. ;
.
15.
The any
If
the diameter through
point on the tangent at
T meet
P
and
to a parabola
the curve in Q, then
nP- = 4:SP.RQ.
PQ
the chord
If
IG.
subtends a right angle at If
17.
PN and XX'
the circle on
which aS',
then
normal at
is
P
to a parabola
SQ = 2.S'P.
P'X' be two ordiuates
of a parabola such that
as diameter touches the parabola, then
XP + X'P' = XX'. 7'P and
IS.
QL
are
TQ
are two tangents to a parabola, and
drawn perpendicular
respectively to
TQ and
Tl',
PK
and
prove that
STK and STL are equal. PP and PP' are tangents to a paral)ola, and the diameter through T cuts the curve in Q. If PQ, P'Q cut TP', TP respectively in R and A", and the diameters through li and R' cut the curve in V, V respectively, prove that PV, P'V intersect on I'Q.
the triangles 19.
[Project the parabola into a circle to
PP'
and the
line
through
7'
parallel
to infinity.]
Pi"Ove that no circle described on a chord of a parabola as 20. diameter can meet the directrix unless the chord be a focal chord,
and then the
chord
circle touches the directrix.
If 7'P
21.
PQ
22.
and
TQ
be a pair of tangents to a parabola and the
be normal at P, then
The
triangle
ABC
focus, prove tliat the lines
SC
TP
bisected by the directrix.
through A, B,
C
S
as
perpendicular to 8A, SB,
respectively are concurrent. 23.
The tangents
to a parabola at
R
and R'
.
Prove that
PR
P
and P'
intersect in
SPQ, SPQ' meet the and QR' are parallel.
circles circumscribing the triangles
in
is
circumscribes a parabola having
Q
;
the
axis again
THE PARABOLA
142
A given
24.
triangle
ABC
moves
in a plane, with one side
passing through a fixed point, and with the vertex
Shew
straight line.
Shew
25.
that the side
AG
will
envelop a parabola.
a line which moves so as to
that the envelope of
two given two given circles
intercept equal chords on
circles is a
radical axis of the
as the tangent at its vertex.
A
2G. axis.
line
xiQ
meets a parabola
drawn
is
parallel to
Prove that the ordinate P and J)-
of
Q
is
P
in
AB
on a given
^4
Pp
to
parabola having the
and p on the same side of the meet the curve again in Q.
equal to the
sum
of the ordinates
of
The distance
27.
of the point of intersection of
a parabola from the axis
is
half the
sum
two tangents
to
of the ordinates of their
points of contact.
LL' be the latus rectum of a parabola and the tangent
If
28.
P meet
at any point
A
29.
cii'cle
Shew
the focus S.
that at
L
QQ'
is
SL LP = VL .
.
VL'.
that the parabola meets the circle again or not
accoi'ding as the latus rectum 30.
in F, then
touches a parabola at a point P, and passes through
is
the normal at
(^
or
is
not
less
than SP.
to a parabola meeting the parabola
again in Q', QP is equally inclined to the axis with the normal and T is the middle point of QQ' and meets the curve again in P ,
;
PV meets
the axis in
R
;
shew that
QSPP
lie
on a
circle,
*S'
being
the focus.
Two paralwlas with equal latus rectum are on the same and are such that the part of any tangent to one which is cut off by the other is equal to the perpendicular upon this tangent from the focus of the first parabola. Shew that the latus rectum 31.
axis,
of each
is
32.
sixty-four times the distance between the vertices.
A
circle
a parabola at both ends of a double
touches
The normal at P meets the circle in and the parabola in Q. The diameter of the parabola through meets PP' in U. Prove that the circle Q RU iowches PP' at U.
ordinate
PP'
to the axis.
R Q
EF is
a double ordinate of the axis of a parabola, R any and the diameter through R meets the curve in P the and the diameters through E and F. tangent at P intersects in and FN. Prove that PR is a mean proportional between 33.
point on
it,
;
M
N
EM
34.
If a parabola roll
being originally in contact, the fixed parabola.
on another equal parabola, the vertices its
focus will trace out the directrix of
THE PARABOLA
143
35. If P be any point on a parabola whose vertex is A, and PR perpendicular to AP meet the axis R, a circle whose centre is R and radius RP will pass through the ends of the ordinate to the
Also if common tangents be drawn to the and parabola the ordinates at the points where they touch the
parabola through R. circle
parabola will be tangents to the 36.
circle.
Through any point on a parabola two chords are drawn
equally inclined to the tangent there.
Shew
that their lengths are
proportional to the portions of their diameters bisected between
them and the
Y in of
curve.
TQ, TR, tangents to a parabola, meet the tangent at P in 37. and Z, and TU is drawn parallel to the axis, meeting the parabola Prove that the tangent at U passes through the middle point U. YZ, and that, if S be the focus,"
YZ- = ^HP.TV. 38.
PQ
a normal chord of a parabola meeting the axis in G.
is
Prove that the distance of G from the vertex, the ordinates and Q, and the latus rectum are four proportionals. 39.
it
intersection
at a constant angle.
a straight
is
The radius
The diameter
Prove that the locus of their
line.
curvature at an extremity of the latus
of
rectum of a parabola 41.
P
Lines are drawn through the focus of a parabola to cut the
tangents to
•40.
of
is
equal to twice the normal.
at either extremity of the latus rectum of a
parabola passes through the centre of
curvature
at
its
other
extremity. 42.
in
PQ
If the tangent at
and the = iPT. T,
43.
If
R
on a parabola, 44.
45.
of
any point
/* of a parabola meet the axis curvature meet the curve in (?, then
he the middle point of the radius of curvature at
PR
P
subtends a right angle at the focus.
The tangent from any point
curvature at
Pof
circle
its
vertex
The chord
is
of curvature
iPYa i^arabola
is
of a parabola to the circle of equal to the abscissa of the point.
,
through the vertex
Y being
the focus on the tangent at
/-•.
A
at
any point
the foot of the perpendicular from
144
CHAPTER
XII
THE ELLIPSE
We
137.
have already in
^§
98 and 99 indicated the general
shewing that it has two axes of symmetry at right angles to one another and intersecting in C the centre. On the major axis AA' are two foci S and ^*', at a distance equal form of the
to
CA
foci it
ellipse,
B
from
and B' the ends
of the
minor
correspond directrices at right angles to
X
externally in
eccentricity.
major
and X'
so that
axis,
A A'
and to these and cutting
GS CA = CA CX =
It is convenient to call
:
:
A
and
the
e,
A', the ends of the
axis, the vertices of the ellipse.
It will
be understood that the smaller
is
the ratio
CS CA :
the more does the ellipse approximate to circular form, and the greater
C>S'
:
CA
is
without reaching unity the more does the
ellipse flatten out.
We
have always
CS'=CA'-CB' so that, keeping CA constant, CB diminishes as GS increases, and vice versa. And we have already explained that a circle may be regarded as the limiting case of an ellipse whose two foci coincide
We which
with the centre.
now proceed
all ellipses
Sum
138.
of focal distances constant.
The sum of the focal distances of any point on an constant and equal to AA'.
Prop. ellipse is
to establish the chief geometrical properties
have in common.
THE ELLIPSE Let through
145
P be any point on an ellipse, MPM' P to the directrices as in the figure.
Then
>S'P = e PM and S'P = e PM' SP + S'P = e (MP + PM) = e XX = 2e.CX = '2CA = AA'. .
. • .
.
;
.
Two
Cor.
the perpendicular
confocal ellipses (that
is,
which have both
foci in
conunon) cannot intersect. just proved shews that an ellipse
The proposition
139.
distances from two fixed points in the plane
we
learn that an ellipse can be
is
sum
may
of
whose
constant.
And
be regarded as the locus in a plane of a point the
drawn by tying the ends
of
a
piece of string to two pins stuck in the paper so that the string is
not tight, and then holding the string tight by means of the
pencil pressed against to
make
its
mark
By keeping
it,
and allowing the point of the pencil
in all possible positions thus determined.
the string the same length and changing the
distance between the pins
same major
we can draw
ellipses all
axis but having different eccentricities.
having the It will
be
seen that the nearer the pins are together the more does the ellipse
approximate
140.
Prop. bisect
to circular form.
Tangent and Normal. The tangent and normal
respectively the exterior
and
at
any point of an
ellipse
interior angles between the
focal distances of the point. A. G.
10
THE ELLIPSE
146
T
Let the tangent and normal at and G respectively.
P
meet the major axis
Then hy^lll SG = e.SP, and S'G = .-.
.-.
.-.
e
.
in
8' P.
SG:GS' = SP:PS';
PG bisects PT which
the angle
SPS'
at right angles to
is
PG
must
bisect the
exterior angle of SPS'.
CG.CT=CS-'
Cor. for since
SP
PG
and
PT
are the bisectors of the angles
between
and S'P, (GT, SS')
= - 1.
Prop.
If SY, S'Y' be the perpendiculars from the foci on any point P of an ellipse, Y and Y' lie on the described on the major axis A A' as diameter, and
141.
the tangent at circle
SY.S'Y' = BC\ Produce
SY
to
meet S'P
in
K.
ASPY= AKPY ^SPY=ZKPY (§140) Z SYP = 4 KYP being right angles and PY is
Then for
.-.
. •
.
PK^SPsiUclKY^^SY, KS'= KP + PS' = SP + PS' = AA'
Now
CY is
since
Y and C
parallel to
(§
138).
are the middle points of
S'K and CY = ^S'K= CA.
common.
SK
and SS',
THE ELLIPSE
Y
Thus
(and similarly Y')
lies
147
on the
circle
AA'
on
as
diameter.
Moreover as Y' YS again in a point
Z
Y'Z goes through
is
a right angle,
meet the
will
circle is,
C.
And ACSZ = ACS'Y' .-.
YS
such that Y'Z will be a diameter, that
so that
SZ=S'Y'.
SY.S'Y'^SY.SZ = AS SA' = CA' .
CS'^
= BC\
P will 1. The SP and S'P in points £'aild E' such that PE = PE'= AC. PE'==CY = AC and For PYGE' is a parallelogram. diameter parallel to the tangent at
Cor.
meet
.-.
PE = AC.
similarly
Cor.
2.
The envelope
perpendicular on
S
which has Cor.
3.
it
within
it, is
side of
an
The envelope
the perpendiculars on
same
it,
is
of a line such that the toot of the
from a fixed point S
it
constant
ellipse
lies
having
*S'
on a fixed for
circle,
a focus.
of a line such that the product of
ft-om is
an
two fixed points, lying on the ellipse having the fixed points
for its foci.
Def. is
The circk on the major
axis of an ellipse as diameter
called the auxiliary circle.
142.
Prop.
If TQ and TQ' be a pair of tangents to an and CT meet QQ' in V and the curve
ellipse wJiose centre is C,
inP,CV.CT=CP\ 10—2
THE ELLIPSE
148
For let PC meet the ellipse again in QQ' are pole and polar, {TV, PF) = - 1. .-.as
143.
C
is
the middle point ofPP',
The preceding proposition
F'.
Then
as
T and
GV CT=GP\
is
.
an important one and
includes the following as a special case
If the tangent at P meet the major and minor axes in and PN, PM he the ordinates to these axes then
GN .CT^GA' GM.Gt=GB\
p
T and
t,
THE ELLIPSE 144.
Prop.
149
normal at any point P on an ellipse G and g, and the diameter tangent at F in F, then FF.FG=BC- and
If
the
meet the major and minor axes in jmrallel
to
the
FF.Fg = AC\ Draw
the ordinates
meet the diameter
FX and FM to
the axes and let these
parallel to the tangent at
in the figure. t
F
F
in
K and L as
THE ELLIPSE
150
Pair of tangents.
144b.
Prop. an
ellipse
Let
The two tangents chawn from an external point to angles with the focal distances of the point
make equal
TP
and
TQ
be the tangents
;
it
is
required to prove
/.PTS = ^S'TQ. Draw 8Y,
S' Y' perpendicular to
TP, and
8Z
and S'Z' per-
pendicular to TQ.
Then SY. S'Y' = BC'^SZ. S'Z' .-.
(|
141)
SY:8Z = S'Z':S'Y'.
= supplement of Z YTZ (since SYTZ is cyclic) = z Y'S'Z' (since Y'TZ'S' is cyclic). A s SYZand S'Z' F'are similar.and Z SZY= Z S'Y'Z',
Also Z F5f^
.
the
•.
But Z ;SfZF= Z *STF in the same segment and Z S'T'Z' = Z >S^TZ' in the same segment. .-.
145.
Prop.
ZSTY=S'TZ'.
Director Circle. The locus of points, from which the tangents to an a ciixle {called the director circle
ellipse are at 7'ight angles is
of the
ellipse).
TP and TQ be two tangents at right angles. SY perpendicular to TP to meet S'P in K. Then by § 141, SY= YK and S'K = AA'. Let
Draw
THE ELLIPSE Also
AaSFT= A A^FT,
Y are
the angles at .
•.
>ST =
for
151
SY= KY and YT is common and
right angles.
KT and z KTP = z STP = z QT8' Z ATra' = Z
.-.
PTQ = a
(§
144 b).
right angle.
T
Now 2CT= +
26'»S'-^
=
.ST-
+ ^T-
(§
10)
= 4C^^ .
CT' = 2CA' - CS' = 2CA ' - {CA' - CB') = CA' + GB\
•.
Thus the locus of T is CA- + CB'-.
a circle round C, the stjuare of whose
is
radius
146.
Conjugate Diameters.
The student lines of a conic,
is
already familiar with the idea of conjugate
two
lines
being called conjugate when each
contains the pole of the other.
meet
When
a pair of conjugate lines
in the centre of an ellipse, each being a diameter it is
convenient to
call
them
conjugate diameters.
It
is
clear that
they are such that the tangents at the points where either
meets the curve are parallel to the other.
Moreover
all
the
chords which are parallel to one of two conjugate diameters are bisected
by the other
(§ 95),
and these chords are double
ordinates of the diameter which thus bisects them.
The axes
of the ellipse are that particular pair of conjugate diameters
which are mutually at right angles.
THE ELLIPSE
152 147.
of an
Prop. If QV he an ordinate of the diameter and BCD' be the diameter conjugate to CP,
PGP'
ellipse,
QV'-:PV.VP'=CIP:CP\
Fur, producing
QV
to
meet the
ellipse
again in
Q',
by
Newton's theorem we have
VQ VQ' VP. VP' = CD CD' CP CP' VQ' = - VQ, CD'=- CD, CP' = - CP; QV':PV.VP' = CD':CP\ .
But
:
.
:
.
.-.
148.
If
Special cases of the preceding proposition are these
PN and PM he
ordinates of the major
ellipse then
PN' -.AN. NA' = BC A C' PM-':BM.MB' = AC-':BC\ :
B P,--""^
:
and minor axes of an
THE ELLIPSE For
SL
if
be the semi-latus rectum
SD AS
.
:
SA' - BG^ :AC' and
These properties
149.
153
in §
AS
.
SA' = BC\
148 shew that an
ellipse
may be
regarded as the locus of a point in a plane such that the square of
its
distance from a fixed line
in the plane bears a constant
I
ratio to the product of its distances I'
and
from two other fixed lines
perpendicular to the former and on opposite sides of
I",
the point.
The I" are
line
I
is
one of the axes of the
ellipse,
and the
lines
I'
and
the tangents at the ends of the other axis.
The property
established in
147 shews that an
§
ellipse ma}-
also be regarded as the locus of a point in a plane such that the
square of
its
distance from a fixed line
constant ratio to the product of fixed lines
I'
and I" which are
necessarily perpendicular to point.
and
The
line
I" are the
I
is
its
I
in the plane, bears a
distances from two other
parallel to each other (but not
I),
and on opposite
sides of
the
a diameter of the ellipse and the lines
tangents at the points where
For the student can
I
l'
meets the curve.
easily prove for himself that in the
notation of § 147
QV^ :PV. VP' = :
square of perpendicular from
product of perpendiculars from
Q
on tangents at
Auxiliary Circle. Prop. If P he any point on an 150. ordinate of the major axis, and if NP meet
ellipse
Q on PP' Q and Q'
and
PN
the
the auxiliary circle
in p, then
NP:Np = BC:AC. For by
§
148
PN':AN.NA' = BC-':AC-' and as Z ApA' being
in a semicircle is a right angle
pN^.-.
= AN.NA']
PN:pN=BG:AC.
P and p are said to be corresponding points on the ellipse and the auxiliary circle. The tangents at two corresponding
THE ELLIPSE
164 points will
For (§
let
meet the
line of the
the tangent at
P
major axis in the same point.
meet
it
in
then
T,
CN'.CT=CA-
143). .'.
7' is
the pole of i^N for the
p
goes through T.
if
an ordinate
The student can prove
for
circle,
that
is,
the tangent at
himself by the same method that
FM to the minor axis
meet the
circle
on BB' as
diameter in p' then
PM:p'M=AC:BG. From
all
this
it
follows that if the ordinates of a diameter
of a circle be all divided in the
same
ratio,
the points of division
trace out an ellipse having the diameter of the circle as one of its axes.
Prop. If GP and CD he a pair of conjugate semi151. diameters of an ellipse, and p, d the points on the auxiliary circle corresponding to and D, then pGd is a right angle.
P
Let the tangents at
Draw
the ordinates
Now
since
CD is parallel .-.
But
as
P
and p meet the major axis
in
T.
PN, DM. to
TP,
/\PNT is similar to ADMC.
PN:DM = NT:MC.
PN :pN = BG:AG = DM: dM, PN:DM = pN:dM. .-.
THE ELLIPSE
pN:dM==NT:MC,
.-.
that
pN NT = dM
is
.'.
:
as the
As pNT
155
unci
dMC
:
MC. and
i\^
be conjurjate semidiameters
of
have the angles at
ilf
equal, they are similar. .-.
.'.
Cd
is
parallel to Tp.
Z dCp = Z CpT = a right
• .
zMCd^zXTp.
.
angle.
pX = CM and dM = CN for A CNp = A dMC.
Cor.
Whence
also
we have
PN:CM = BC:AC, DM:CN = BC.AC. Prop.
152.
an
If
CP
and
CD
ellipse
CP-'+CD"-
For using the figure of the
cp^ and CD'
.-.
=
CM-'
=
GX-'
last proposition
+ PX' =
+ DM' = i^^' +
CP' + CD' = (l+
= CA-'+CB\
av-^
+
~,
.
px-'
RC-
7?r"
1^ ^^' " ^^' I^ ^'^'• "^
1^) (pX' + CX^)
= (i+^)ac-' = ac' + bc"^.
THE ELLIPSE
15G
Thus the sum an
of the squares of two conjugate diameters of
ellipse is constant
Or we may prove the In
§
= AA"^ + BB'".
and
proposition thus
151 we proved
pN = CM,
CM' +
= pN'- +
. • .
CN-'
CN-'
:
= Cp' = A C\
In exactly the same way by drawing ordinates to the minor axis
and working with the
FN' + Whence by
DM''
on BB' as diameter we have
= BC\
GP' + CD-
= AC + BG\
Equiconjugate Diameters.
152a.
There
addition.
circle
is
one pair of conj ugate diameters of an ellipse which
are equal to one another namely those which
lie
along the
formed by the tangents at the extremities of the major and minor axes. diagonals of the
rectangle
X
157
THE ELLIPSE Since
is
the middle point of SS'
= (ST + spy - 2SP S'P = 4^CA'-2SP.S'P. .
.-.
.
Prop.
153.
twrmal
SP S'P = 2CA' - CP' - CS' = CA"- + CB' - CP' = CD' (by §152).
at
P meet
// P be amj point on an BCD', the diameter conjugate
ellipse, to
CP,
and
the
in F, then
PF,CD = AC.BC. Draw
the tangent at
S'Y' from the
Join
SP
foci (jn
Pand
(h-opthe perpendiculars .S'Fand
it.
and S'P, and
let
S'P cut DD'
Then the As SPY, S'PY'
are similar.
in E.
THE ELLIPSE
158 .-.
.-.
SY:8P==S'Y':S'F
SV. S'Y' SP S'P = SY' SP' = PF' PE' :
:
.
:
As SYP,
since the
PFE are .-.
that
is,
The area
Cor.
similar.
BC'-:CD-' = PF':Aa' PF.CD = AG.BC. of the parallelogram formed by the tangents
at the extremities of a pair of conjugate diameters
is
constant
= 4.AC.Ba
= 4 area of parallelogram PD^ = ^PF.GD = 4:AG.BG.
For the area
Circle of Curvature. 154. 'point
P
Prop. The chord of the circle of curvature at any of an ellipse and through the centre of the ellipse is
2G]>
GP Let
Q be
•
a point on the ellipse near to
P
and
QV
the
ordinate of the diameter PGP'.
Consider the in Q.
QK
Let
circle
QK be
touching the ellipse at
P
and cutting it Let
the chord of this circle parallel to GP.
meet the tangent
at
P
in R.
THE ELLIPSE
Then from
159
the circle
RQ.RK = RP\
Thus the chord being the limit of
of the circle of curvature through the centre
RK when
Q
approaches
X Limit
F
VF'
2CD' ^x2Crp=^^. Cor.
F
The diameter
of the
circle
of curvature
2CD"-
being the point in which the normal meets CD, for
Diameter
2011 =
CF
sec
(
Z between normal and OF)
CF FF. :
Diameter
2CJy
2CD'
FF
ACTBC'
FF
THE ELLIPSE
160
EXERCISES Prove that in the notation
1.
as SL
If
2.
be
tlie
:
of this chapter
ex = CS'^
:
CA^
semi-latus rectum of an ellipse then
SL =
e
.
SX
;
prove from this that
Obtain also the length of the latus rectum by using the fact 116, 117) that the lengths of
(§§
two
focal chords are in the ratio
of the squares of the diameters parallel to then).
If Y,
3.
an
ellipse
Z be
the feet of the perpendiculars fi'om the foci of is the ordinate ; prove
on the tangent at P, of which
PN
YNZ passes
that the circle circumscribing
through the centre of the
ellipse.
circle
If P be any point on an ellipse whose foci are S and S' the circumscribing SPS' will cut the minor axis in the points
where
it is
\.
met by the tangent and normal at two
/^
touch internally the locus of the centres of circles touching them both is an ellipse, whose foci are the centres of the given circles.
and
circles
5.
If
6.
If the tangent at
NG be
P
to
an
ellipse
meet the major axis in
'l\
the subnormal,
CT .NG=BC\ 7.
If
PN
be the ordinate of any point
Y, Y' the feet of the perpendiculars
P, then
PN bisects the angle
from the
P foci
an ellipse and on the tangent at
of
YNY'.
normal at P to an ellipse meet the minor axis in
If the
:
9. If the normal at P meet the major axis in G, PG is a harmonic mean between the perpendiculars from tlie foci on the tangent at P.
10.
If
an
ellipse inscribed in a triangle
have one focus at the
orthocentre, the other focus will be at the circumcentre.
THE ELLIPSE 11.
an
If
the locus of
ellipse slide
its
centre
is
161
between two straight
lines at right angles,
a circle.
Lines are di'awn through a focus of an ellipse to meet the
12.
tangents to the ellipse at a constant angle, prove that the locus of the points in which they meet the tangents
The
13.
is
a
circle.
locus of the incentre of the ti'iangle
the foci of an ellipse and any point on the curve
The
14.
ellipse
whose vertices are an ellipse.
is
opposite sides of a quadrilateral described about an
subtend supplemeutaiy angles at either focus.
Prove that the foci of an ellipse and the points where any it meets the tangents at its vertices are concyclic.
15.
tangent to
If
16.
CQ
be a semidiameter of an ellipse conjugate to a chord
which is normal normal at Q. 17.
If
P
to the
curve at P, then
CP
be any point on an ellipse, foci
vertex, then the bisectors of the angles
.S'
is
conjugate to the
and
.S",
and
.4
be a
PSA, PS' A meet on the
tangent at P. 18. In an ellipse whose centre is C and fuci S and S', GL is drawn perpendicular to C P, and CJ/is drawn parallel to S' P meeting PG in M. Prove that the triangles CLM, CMP are similar. 19. A circle is drawn touching an ellipse at two points, and Q any point on the ellipse. Prove that if ^7' be a tangent to the circle from Q, and QL perpendicular to the common chord, then
is
QT=^e.QL. If a
20.
parabola have
of an ellipse, and touch ellipse
and parabola
21.
Shew how
axes of an
ellipse,
will
its
focus coincident with one of the foci
minor axis, a common tangent to the subtend a right angle at the focus. its
to determine the
magnitude and position
of the
having given two conjugate diameters in magni-
tude and position. 22.
Construct an ellipse when the position of
self -conjugate triangle
its
centre and a
are given.
If P be any point on an 23. and AP, A' P meet a directrix in
ellipse
whose vertices are A and A
E and
F, then ^i*^ subtends a right
',
angle at the corresponding focus. 24.
Deduce from Ex. 23 the property that PX'^
:
AN
.
NA'
constant. A. G.
11
is
THE ELLIPSE
162
Prove that chords joining any point on an ellipse to the 25. ends of a diameter are parallel to a pair of conjugate diameters.
[Two such chords are
ends
called supplemental chords.]
and intersect it at the and PQ' are fixed in direction.
If a circle touch a fixed ellipse at P,
26.
diameter QQ', then
of a
Ellipses
27.
PQ
have a common fixed focus and touch two fixed
straight lines, prove that their director circles are coaxal.
SY is
28.
the perpendicular from the focus S of an ellipse on a such that SY = YK. Prove
K the point in 5 F produced
tangent, and
that the square of the tangent from A' to the director circle
is
double
the square on SY.
The
29.
curvature at an extremity of one of the equal
circle of
conjugate diameters of an ellipse meets the ellipse again at the extremity of that diameter.
of
PN be the ordinate of a point P on an ellipse, 26'2)2 x PN. curvature in the direction of PN = .30.
If
31.
If
S and
the chord
and B an extremity of minor axis in the centre
S' be the foci of an ellipse
the minor axis, the circle
SS'B
will cut the
of curvature at B.
The
32.
circle of
curvature at a point
P
of
an
ellipse passes
through the focus S, and SE is drawn parallel to the tangent at P to meet in E the diameter through P ; shew that it divides the diameter in the ratio 3:1.
The
33. ao'ain in
Q
;
circle
which touches the {TO, EF) = -\. 34.
Q and
ellipse
The tangent shew that
Q'
;
P on an ellipse cuts the curve meets the other common tangent, prove and circle at E and F, in
of curvatui-e at
the tangent at
P
;
P to an ellipse meets the equiconjugates CP is a symmedian of the triangle QCQ'.
at
ia
163
CHAPTER
XIII
THE HYPERBOLA 155. is
We
have eeen in
|§ 100,
101 that a hyperbola, which
the projection of a circle cut by the vanishing
line,
axes of symmetry at right angles, one of which,
has two
named the
what are called the vertices and A', while the other called the conjugate axis does not meet the curve. These two axes meet in C the centre of the curve, and there are two tangents from C to the curve having These tangents are called their points of contact at infinity. the asymptotes and they make equal angles with the axes. transverse axis, meets the curve in
A
The curve has two
foci
S and
S' lying on the line of the
transverse axis, and such that the feet of the perpendiculars
from them on the asymptotes diameter.
The
directrices
lie on the circle on which are the polars of the
A A'
as
foci are
and pass through the feet and X' be the points in which cut the transverse axis and C be the centre, then
at right angles to the transverse axis, of these perpendiculars.
the directrices
the eccentricity 156.
(e)
If A'
= CS CA = CA CX. :
:
In this chapter we shall
set
forth
the
principal
common. Some of these and can be established in
properties that all hyperbolas have in
same as those of the ellipse But the fact that the hyperbola has a pair of asymptotes, that is, tangents whose points of contact are at infinity, gives the curve a character and properties of its are the
much
the same way.
own.
11-2
THE HYPERBOLA
164
Difference of focal distances constant.
157.
Prop.
Tlie difference
a hyperbola
constant,
is
of the fucal distances of any point on and equal to the length A A' of the
transverse axis.
Let
P
be any point on the hyperbola.
PMM'
Draw directrices,
as
the
in
figure
perpendicular
to
the
then
and SP = e.PM. S'Pr^SP = e.XX'^AA'.
S'P=e.PM' .-.
For points on the one branch we have S'P on the other SP — S'P = A A'.
— SP =AA' and
for points
Cor.
Two
confocal hyperbolas cannot intersect.
Tangent and Normal. 158.
Prop.
The tangent and normal
at
any point of a
hyperbola bisect respectively the interior and exterior angles of the focal radii of the point.
Let the tangent and normal at in
T and
P
meet the transverse axis
G.
Then by
§
111,
e SP, S'G = e S'P. SG:SP = S'G:S'P.
SG = .-.
.
.
THE HYPEEBOLA .•.
PG
is
165
the bisector of the exterior angle of SPS', and
PT, perpendicular
PG, must
to
therefore bisect the interior
angle.
Cor.
1.
CG.CT= CS\
Cor.
2.
If
an
ellipse
for {SS',
TG) = -
1.
and hyperbola are confocal their
tangents at the points of intersection of the curves are at right angles, or, in other words, the curves cut at right angles.
SY, S'Y' he the perpendiculars from the a hyperbola at any point P, Y and Y' will on A A' as diameter {called the auxiliary circle),
Prop.
159.
//'
foci on the tangent lie
on the circle
and 8Y. S'Y' Let
luill
SY meet
Then
to
he constant.
S'P
since
K.
SPY = A KPY (§ 158) ^SYP=ZKYP
^
and and
in
PF is common, .-.
And
A5rPF= A/fPF
SY^YK, PK = SP.
and since
parallel to
F and C S'K and
are the middle points of *S^
and SS',
CY is
GY= ^S'K = i {S'P - KP) = ^{S'P-SP) = ^AA' = CA. Y (and similarly F') lies on the circle on AA'.
Thus
THE HYPERBOLA
166 If S'Y'
right angle,
meet the circle again in Z, then since YY'Z YZ must be a diameter and pass through C.
^SCY= AS'CZ
Also
and
and S'Z=SY.
constant.
is
Cor.
SP
a
SY.S'Y'=S'Z.S'Y' = S'A'.S'A^CS"--CA'
.-.
which
is
1.
The diameter parallel
to the
tangent at
P will meet
ST in points E and E' such that PE= PE' = GA.
Cor.
The envelope
2.
perpendicular on
which has S outside Cor.
3.
of a line such that the foot of the
from a fixed point
it
it, is
S
lies
a hyperbola having
The envelope
>S'
on a fixed
circle
for a focus.
of a line such that the product of
the perpendiculars on
it
from two fixed points, lying on opposite
constant
is
a hyperbola having the fixed points for
sides of
it, is
its foci.
Compare 160.
On
§
141, Corr. 2 and
3.
the length of the conjugate axis.
We
have seen that the conjugate axis of a hyperbola does not meet the curve, so that we cannot say it has a length in the same way that the minor axis of an ellipse has for its length that portion of
it
intercepted by the curve.
It is convenient,
and
this will
be understood better as wo
THE HYPERBOLA proceed, to measure off a length
that
B
This
will
BG\
BB' on
and B' are equidistant from
BC' -
GS-^
make SY.S'Y'
(Compare
-
GA"-
G,
B
and B' do not 161.
.
A'S.
in the preceding proposition equal to
§ 141.)
The length BB' thus defined
BB'
the conjugate axis such
and
= ^,S'
will for
the length of the conjugate axis, but stood that
16^
is
it
convenience be called
must be
clearl}^
lie
on the curve.
It will easily
be seen that
if
a rectangle be drawn
having a pair of opposite sides along the tangents at
and having
its
under-
not a diameter length of the hyperbola, for
A
and
A',
diagonals along the asymptotes, then the portion
of the conjugate axis intercepted in this rectangle will be this
length
BB' which we have marked
off as
explained above.
meet the asymptote CO in 0, and the directrix corresponding to S meet CO in K, we have, since GK8 is a right angle and GK=GA (§ 101), For
if
the tangent at
A
AGKS= A GAG.
Hence
A0' = SIC-= GS' - GIO = GS' - GA\
THE HYPERBOLA
168
Pair of tangents. 162.
Prop.
TJte
two tangents drawn
from a
point to a
hyperbola make equal or supplementary angles with the focal distances of the point.
Fig.
Let TP, angles
TQ
be the tangents,
STP, S'TQ
perp. to
SY. .-.
Also
it is
required to prove that the
are equal or supplementary.
Draw SY, S'Y' Then
1.
S' Y'
TP, and SZ, S'Z'
= BO' = SZ
.
to
TQ.
S'Z'.
SY:SZ=S'Z':S'Y'.
Z YSZ = /.Z'S'Y' these being the supplements
of the
THE HYPERBOLA
169
ZTY and Z'TY' in fig. 1, and SZTY and 8'Y'TZ'
equal angles
equal to YTZ, since
Hence the As
SYZ and
z >S'rP
while in
2
fig.
it
in
fig.
2 each being
are cyclic.
S'Z'Y' are similar and
= z >SZF= z aS"F'Z' = z = supplement of Z >S"TQ
>S'TZ' in fig.
1,
= Z S'TQ.
Thus the two tangents from an external point make equal supplementary angles with the
focal
or
distances of the point
according as the tangents belong to opposite branches or the
same branch
of the curve.
Director Circle.
163.
Prop.
The locus of points tangents from tvliich to a hyperbola are at right angles is a circle (called the director circle of the hyperbola).
Let
TP
Draw
and
TQ
8Y perp.
Then by
159,
§
Also
Thus
ST= KT .
•.
be two tangents at right angles. to
TP
to
meet S'P
in K.
SY= YK and S'K=AA'. A8YT= £^KYT.
and z
KTY = Z ST Y=
Z KTS' = Z
PTQ = a
Z.
QTS'
right angle.
(§
162).
170
THE HYPERBOLA
The hyperbola and each having
its
foci
conjugate are two distinct curves,
its
and
171
directrices,
nor
they in general
will
have the same eccentricity.
The
foci
A A' and CS:GA. the line
S and
S' of the original hyperbola
are such that CS-
= GA^ + GB-, and
lie
on the
The foci 2 and S' of the conjugate hyperbola of BB' and are such that CI- = GA^ + GB\
Thus 02 = G8, but the
eccentricity is (7S
the same as that of the original hyperbola
:
if
line of
the eccentricity
GB, which GA = GB.
lie
is
is
on
only
In this special case the asymptotes are the diagonals of a square
(§
IGl) and are therefore at right angles.
When
the asymptotes are at
said to be rectangular.
i-ight
angles the hyperbola
is
In the next chapter we shall investigate
the special properties of the rectangular hyperbola.
The conjugate hyperbola
is,
as
we
shall see, a very useful
adjunct to the hyperbola and considerable use will be it in
what
made
of
follows.
Asymptotic properties. Prop. // R be any point on an asymptote of a hyperand RN^ perpendicular to the transverse axis meet the hyperbola in P and 'p then RP Rp = BG^. 166.
bola,
.
THE HYPERBOLA
172
A
Let the tangent at
meet the asymptote on
Avhich
R
lies
in E.
Then
fl
being the point of contact of the asymptote with
the curve at infinity
we have by Newton's theorem
RP Rp .
.'.
:
Rn^'
= EA-
:
En--.
RP.Rp = EA' = BG\
This can also be written
RN' - PN' = BC\ It
will
presently be seen that this proposition
special case of a
167.
Prop.
of any point
P
//
PN
PN"~:AN.A'N^BC':AG\ 166 we have
§
- PN' = BC\ PN^ = RN'-BG\ RN^ BG"- - RN' EA' = GN' GA\ RN"' - BG' BG' = GN-' - GA' GA\ PN"-:BG' = AN.A'N:CA'. PN':AN.A'N = BG'':AG\ RN-'
.-.
:
:
:
.-.
:
.-.
.-.
only a
he the ordinate to the transverse axis
of a Ityperhola
Using the figure of
But
is
more general theorem.
:
THE HYPERBOLA or
we may
173
write this
BC':AC\ This too will be found to be but a special case of a more general theorem.
Comparing ellipse (§ 148)
this property
we
see that
with the corresponding one in the it
was not possible
property for the hyperbola in the same
way
to establish the
as for the ellipse,
because the conjugate axis does not meet the hyperbola.
If from any poixt
Prop.
168.
hyperbola
RPN,
R
in an asymptote of a
RDM he drawn perpendicular to
and conjugate axes to cut the hyperbola and respectively in P and D, then PD is parallel asymptote, and CP, CD are conjugate lines for both and the conjugate hyperbola.
Let its
n
its
to
conjugate the
other
the hypei^hola
and DJ be the points of contact of the hyperbola and
asymptotes at
We
the transverse
first
infinity.
observe that
For drawing the
AB is
lines
parallel to CVl'.
through
A
and
B
perpendicular to
THE HYPERBOLA
174
the axes to meet the asymptotes, as indicated in the figure, in E,
E\ we have
e,
=EA
EB-.BE'
3IN
Also
A B,
parallel to
is
-.Ae.
for
GA CB = GA:AE^ CN RN = CN CM. GA CN = CB CM. RN"^-PN' = BC' ^^^' ' RM'^ - DM' = AC RN"- - PA''' RM' - DM' = CM' CN' :
:
:
.'.
Now and
:
:
'
.-.
:
:
^RN'-.RM'.
RN':RM' = PN':DM'.
.-.
PD is parallel to MN and therefore to CD.'. PD will be bisected by CO in the point T (say). Now DP will meet Gil' at Df, and we have Thus Thus
(DP, Tn') .'.
and
GP
CD
and
But Cfl and
GP
and
its
CO
lines.
being tangents from
C
to both the hyper-
conjugate are the double lines of the involution
its
pencil formed .-.
Cfl'
= -i.
belong to the involution of which
double
CVl' are the
bola and
will
by the pairs of conjugate
and
CD
lines
through
C.
are conjugate lines for both the hyperbola
conjugate.
For this it follows that the tangent at P to the hyperbola is parallel to CD, and the tangent at D to the conjugate hyperbola is
parallel to
169. '
On
GP.
the term conjugate diameters.
If the lines
bola and
its
PC,
PGP' and DCD'
then
of the hyperbolas.
PGP' not,
DC
is
but
in the figure of § 168
meet the hyper-
conjugate again in the points P' and D' respectively, are called conjugate diameters for each
But
it
must be
clearly understood
a diameter of the original hyperbola, whereas it is
a diameter of the conjugate hyperbola.
that
DCD'
is
THE HYPERBOLA
Of two
175
so-called conjugate diameters one is a diameter of
the hyperbola and the other of the conjugate hyperbola.
The
line
in so far as
BCD' it is
is
a diameter even for the original hyperbola
a line through the centre and
system of parallel chords, but
it is
it Avill
bisect a
not a diameter in the sense
represents a length intercepted by the curve on the line, and D' are not on the hyperbola. DGD' does not meet the hyperbola in real points, though of course as the student
that for
it
D
acquainted with Analytical Geometry will know curve in imaginary points, that involve the imaginary quantity
Prop.
170.
is,
it
meets the
points whose coordinates
V— 1.
The tangents at the extremities of a pair of form a parallelogram whose diagonals lie
conjugate diameters
along the asymptotes.
Let
PGP' and BCD'
be the conjugate diameters, as in the
figure.
We
have already proved
The tangents
at
P
and
(§
D
PD
bisected by
CO.
are respectively parallel to
CD
168) that
is
and GP. These tangents then form with having one diagonal along CO.
GP
and
GD
a parallelogram
THE HYPERBOLA
176
Similarly the tangents at P' and D' meet on Cfl, and those at P', i) and P,
D' on CiY.
The portion
Coil.
of the tangent at any point intercepted
between the asymptotes
is
bisected at the point of contact.
LP = L)C=GD' = Pl.
For
The property given
171.
in the Corollary ot §
independently established by projecting into a
may
170 can be
circle,
and we
use the same letters in the projection without confusion.
Let the tangent at
nn'
P
to the circle
meet the vanishing
line
K.
in
C L
a
The
polar of
K goes through G, since that at K goes through P.
G
goes through
K, and the polar of .-.
GP
Let
is
GP
the polar of K.
meet HH' .-. .-.
in F.
(/iP,
.-.
Thus with
P
an') = -
1.
= -i. {KP,Ll) = -\.
C(ifp, aa')
in the hyperbola
and the point at
L
and
/
are harmonically conjugate
infinity along LI.
.-.
LP = Pl.
R
on an asymptote of a Prop. // through any point 172. hyperbola a line he drawn cutting the same branch of the hyperbola in Q and q, then RQ Rq is equal to the square of the semi.
diameter of the conjugate hyperbola parallel
to
RQq.
THE HYPERBOLA
V be the middle point of Qg. CV cut the hj^erbola in P. Then the tangent at P is parallel
177
Let Let
Let
it
Let
meet the asymptotes
CD
L
in
to Qq.
and
I.
be the semi-diameter of the conjugate hyperbola
parallel to Qq.
n
By Newton's theorem we have
RQ.Rq: .-.
Thus
the line
if
the rectangle
R on We may
point
Rn^-
= LP'
:
LQ.\
RQ.Rq = LP^ = CD\ RQq
RQ Rq .
be always drawn
is
in a fixed direction
independent of the position of the
the asymptote. ^vl•ite
the above relation
RV'-QV'^CD'. And
if
RQq meet ...
the other asymptote in r
.-.
and
.'.
Hence any chord
we have
rV2-qV^=CI)\ rV - qV = RV' - QV\
RV=
Vr
RQ = q>:
of a hyperbola
and the length of
its line
intercepted between the asjanptotes have the same middle point. A. G.
12
THE HYPERBOLA
178
Wc
thus have the following remarkable property of the
hyperbola
:
If Rr joining any tivo points on the asymptotes of a Q and q then EQ = qr.
hyperbola cut the curve in 173.
Prop.
// a
line he
asymptote of a hyperbola in
Q and
q then
qR.
to
RQ =
of the liyperbola parallel
drawn through a point
R
on an
meet opposite branches of the curve CP'- where GP is the semi- diameter
to
Qq.
For by Newton's theorem
= GP GP' CHl RQ.Rq = -GP\
RQ.Rq: .-.
.-.
As
RVt"-
it
:
qR.RQ = GP\
in the preceding article
portion of
.
we can shew
that
Qq and the
intercepted between the asymptotes have the same
middle point.
Prop. If QV he an ordinate of the diameter PGP' 174. and DGD' the diameter conjugate to PP' then
QV-':PV.P'V = GD':GP\ Let Q V meet the asymptote Gfl in
R
and the curve again
in Q'.
Through PGP'.
R
draw the chord
qq' of the hyperbola parallel to
THE HYPERBOLA
179
Then by Newton's theorem
VQ VQ' VP. VP' = RQ. RQ' Rq - QV VP VP' = CD' - CP\ .
:
.-.
that
:
:
.
Rq'
:
.
is
QV':PV.P'V=CD':CP"-.
This
the general theorem of which that of
is
§
107
is
a
special case.
We may
write the relation as
QV:CV'-CP'=CD':CP\ 175.
From
§§
1G7, 174
we can
see that a hyperbola
may be
regarded as the locus of a point in a plane such that the ratio of the scjuare of
product of
its
its
distance froni a fixed line
I
varies as the
distances from two other fixed lines
parallel to one another
and such that the point
is
l'
and
I"
on the same
side of both of them.
If
I'
and
l"
be perpendicular to
axis of the hyperbola
If r
and
/"
l',
I"
I
then
and
l',
I"
/,
the transverse
is
I
the tangents at
are not perpendicular to
of the hyperbola, it
and
then
its vertices.
I
is
a diameter
are the tangents at the points
where
meets the hyperbola.
12—2
THE HYPERBOLA
180
Prop.
176.
of
tJie
If QQ', RR'
squares of
tlie
he chords of
a hyperbola
OQ OQ' OR OR'
then the ratio
secting in
.
:
diameters -parallel
.
is
equal
to the respective
inter-
to that
chords.
Let OQQ' meet an asymptote in L.
Through r
and
L
draw irr'
parallel to
ORR'
to
meet the curve
in
r'.
Then by Newton's theorem
OQ OQ' lOR.OR' =LQ.LQ':Lr. .
= sq.
of diameter parallel to
sq. of
:
Lr'
diameter parallel to
This proposition holds equally well
177.
either or both of the chords
hyperbola, provided that
OQ
lie .
if
QQ'
RR\
the ends of
on opposite branches of the
OQ' and
OR
.
OR' be regarded
simply as positive magnitudes. Q, Q' lie
on opposite branches and R, R' on and LQ' are in opposite directions the application of Newton's theorem say
For suppose that
the same branch, then as
we must
LQ so that,
.
for
LQ
LQ' = — QL LQ' = — sq. of diameter asOQ.OQ' = -QO.OQ' we have .
parallel to QQ',
QO OQ' OR OR' = sq.
of diameter parallel to
QQ'
sq.
of diameter parallel to
RR'.
.
:
.
:
THE HYPERBOLA 178.
It
may perhaps seem
181
imnecessarj^ to
make a
separate
proof for the hyperbola of the proposition proved generally for
the central conies in
§
But
117
§
diameters of the curve true.
But our purpose has been to bringthe diameters must be length
117.
out the fact that in
and
itself,
for these the proposition is
as diameters of the hyperbola do not
curve in real points, we wanted to shew
may be used
the conjugate diameter
OQ OQ' and OR OR'
signs of
are different, this
all
meet the
the diameters of
instead.
Whenever the
in the notation of §§ 176, 177
.
.
how
means that the diameters parallel to QQ', RR' them meets the hyperbola. The
are such that only one of
other meets the conjugate hyperbola.
We
179.
can see
now
that
DCD'
if
be a diameter of the
conjugate hyperbola, the imaginary points
8,
S'
in
which
it
meets the original hyperbola, are given by (78^
and
=
Ch'-'
= - CD\
this to the student acquainted with Analytical
also clear
Geometry
is
from the following
The equation
of the hyperbola
is
,.,
.,
a-
=
l
(1), ^
b-
and of the conjugate hyperbola 1
Thus Ci^Tcsponding {ix, ii/)
to every point {x, y)
(2).
on (2) there
is
a point
on (1) and vice versa.
And
if
bola (2) in
180.
a line through the centre meet the conjugate hyper(x,
y)
Prop.
it will
If
meet the
CP
and
original hyperbola in {ix, iy).
CD
be conjugate semi-diameters
of a ItyperboUt
CP' - CD' = CA' - CB\
Draw
the ordinates
conjugate axes.
PN
and
DM
to
the transverse and
THE HYPER150LA
182
These intersect in a point
R
on an asymptote
(|
we have CT- =
C'iV-
+ PN- = CR' - {RN^ - PN^)
^CR'-BC^
(§166)
168),
and
THE HYPERBOLA 182.
Prop.
If
GP
and
CD
of a hyperbola, and the normal at
Draw
the
183
he conjugate semi-diameters
P
meet
CD
PF.CB = AC.BC. perpendiculars SY and S'Y'
in F, then
from the
tangent at P.
Then
that
is
the
As SPY, S'PY being
SY
S'Y'
SP
S'P
S'Y'-SY S'P-
similar
-IPF
we have
PF
foci
on the
THE HYPERBOLA
184
angular points of the parallelogram formed by the tangents at
P,P',D,D'
(§170).
Moreover /\CLl
gram formed by
one quarter of the area of the parallelo-
is
these tangents, that
A GLl = CA which
is
Cor.
.
182),
constant
The envelope
of a line which forms with two fixed
lines a triangle of constant area is a
lines for its asymptotes, its
is (§
GB,
hyperbola having the fixed
and the point of contact of the
line
with
envelope will be the middle point of the portion intercepted
between the fixed
Prop.
184.
lines.
TQ and TQ' he tangents to the same branch GT meet the cu7've in P and QQ' in V, then GV.GT = GP\
//
of a hyperbola, and
This follows at once from the harmonic property of the pole
and
polar, for
we have
{PP\ TV) = - 1. .-.
GV.GT = GP\
Prop. If TQ and TQ' be tangents to o-p-^ositebranches GT meet QQ' in V and the conjugate hyper-
185.
of a hyperbola, and bola in
P
then
VG.CT=GP\
THE HYPERBOLA
185
This can be surmised from the preceding proposition, for
CT
meet the
have
(§
original hyperbohi, in the imaginary point p,
if
we
179)
Cp'=-CP\ .'.
CV.GT^Cj)'
(§184)
= - CP\
VG.CT = CP\
.-.
We give however the foHowing i)urely geometrical which does not introduce imaginary points. Let
BCD' be
the diameter conjugate to
the hyperbola in D, D' and ,
Draw
the ordinate
el to
TQ
QW to
in
PP' and meeting
t.
the diameter DI)', that
is,
PP'.
Then by
similar
As
tWQ.,
tCT
TC:WQ=Gt:tM\ .'.
TC.
]VQ:WQ'=Ct.CW:GW.tW ^Gt.GW-.GW'-Gt.GW.
Butby §184, C'<.(71^=Ci)^ .-.
TG.
WQ WQ' = GD'^ :
proof,
:
CW'^
- GD\
QW
is
THE HYPERBOLA
186
GP'
But
.-.
.-.
The
186.
:
WQ'=CD'
:
GW - GD'^
(§
174).
TG.WQ = GP\ VG.GT=GP\
following arc special cases of the two preceding
propositions.
If the tangent at conjugate axes in to these
P
T and
to t
a hyperbola meet
respectively
the transverse
and PN,
PM he
and
ordiiiates
axes
GT.GN^GA"-
MG.Gt=GB\ For the tangents from Twill be TP and TP' where P' is the PN again meets the hyperbola and the tangents from t will be tP, tQ where Q is the point in which again meets the hyperbola.
point in which
;
PM
187.
Prop. If the normal at P to a hyperbola meet the and conjugate axes in G and g, and the diameter
transverse
parallel to the tangent at
P
in F, then
FP.PG = BG% PF.Pg = AG'.
THE HYPERBOLA This
proposition
can
be
established
corresponding one in the ellipse
9
M
(§ 144).
18^
exactly
like
the
THE HYPERBOLA
188 3.
any point
the tangent at
If
T and SP
asymptote in
/*
\^TP and TO. subtend equal angles at 4.
Shew
that
a
of
hyperbola cut an
cut the same asymptote iu
when a
(^)
then
SQ =
Q'f.
S.'\
pair of conjugate diameters of a hyper-
bola are given in magnitude
and position the asymptotes are Hence shew that there are only two
completely determined.
hyperbolas liaving a given pair of conjugate diameters. 5.
If
two hyperbolas have the same asymptotes a chord
touching the other 6.
If
is
PK be drawn
PH,
parallel to the asymptotes
CO in // and Pll.PK^\CH\
a hyperbola to meet CO' and
[Use
of
one
bisected at the point of contact.
A',
CO, CO' of
then
§ 183.]
The tangent to a hyperbola at P meets an asymptote in T and TQ is drawn parallel to the other asymptote to meet the curve in Q. PQ meets the asymptote in L and M. Prove that LM is trisected at P and Q. 7.
8. From any point R on an asymptote of a hyperbola IIPN is drawn perpendicular to the transverse axis to cut the curve in P RK is drawn at right angles to CR to meet the transverse axis in K.
Prove that
PK is the CN=
[Prove that
normal at P. e'-
.
OK.
%
188.]
Prove that in any central conic if the normal at P meet the axes in G and g then PG Py - CD" where CD is conjugate to CP. 9.
.
10.
If
the tangent at a point
P
of
a hyperbola
meet the
asymptotes in L and I, and the normal at P meet the axes in G and g, then L, I, G, g lie on a circle which passes through the centre of the hyperbola. 11. The intercept of any tangent to a hyperbola between the asymptotes subtends at the further focus an angle equal to half the angle between them.
Given a focus of an ellipse and two points on the curve 12. shew that the other focus describes a hyperbola. 13.
If
P
bfi
any point on a central conic whose
foci are
S and
the circles on SP, S'P as diameters touch the auxiliary circle and have for their radical axis the ordinate of P. S',
The pole of the tangent at any point /* of a central conic 14. with respect to the auxiliary circle lies on the ordinate of P.
THE HYPERBOLA
189
15. If PP' and DD' be conjugate diameters of a hyperbola and Q any point on the curve then QP- + QP'- exceeds QD'~ + QD'- by a
constant quantity. 16.
Given two points
of a parabola
prove that the locus of
axis,
17.
If
its
focus
two tangents be drawn
is
and the direction
of its
a hyperbola.
a hyperbola the lines joining
to
their intersections with the asymptotes will be parallel. 18. If from a point /* in a hyperbola PK be drawn parallel to an asymptote to meet a directrix in K, and S be the corresponding focus, then PK = SP. 19.
If the tangent
meet the axis 20.
in
and normal at a point P of a central conic and PN be the ordinate, XG CT= BC-.
T and G
The base
.
of a triangle being given
contact with the base of the inscribed is
circle,
and also the point
of
the locus of the vertex
a hyperbola.
If tangents be drawn to a series of confocal hyperbolas the 21. normals at their points of contact will all pass through a fi.xecl point, and the points of contact will lie on a circle. 22.
A
hyperbola
is
described touching the principal axes of a
hyperbola at one of their extremities
;
prove tha<^ one asymptote
and that the other the chords of the parabola bisected by the first. parallel to the axis of the parabola
23.
P,
If
an
ellipse
and a hyperbola confocal with
is
it
is
parallel to
intersect in
the asymptotes of the h3'perbola pass through the points of
intersection of the ordinate of
P
with the auxiliary
circle of the
elli]>so.
Prove that the central distance of the point where a 24. tangent to a liyperbola meets one asymptote varies as the distance, parallel to the transverse axis, of the point of contact from the other asymptote. 25.
Tangents PPR', TQT' are drawn to a hyperbola, P, T R\ T' on the other; shew that the on RT' and R'T as diameters are coaxal with the director
being on one asymptote and circles circle.
From any point P on a given diameter of a hyperbola, two drawn parallel to the asymptotes, and meeting the hyperbola in Q, Q' prove that PQ, PQ' are to one another in 26.
straight lines are
;
a constant
ratio.
THE HYPERBOLA
190
The asymptotes and one point on a hyperbola being given, 27. determine the points in which a given Hne meets the curve.
PN be^the ordinate and PG the
If
28.
a hyperbola wliose centre
asymptotes Jin
mean
L and
L',
is
The tangents
normal and the tangent at
CL
then half the sum of
proportional between
29.
C,
CN
a point P of P intersect the
of
and CL'
is
the
and CO.
to a conic
from any point on the director between every pair of conjugate
circle are the bisectors of the angles
lines
through the point.
30.
Given a
focus,
locus of the centre 31.
P
If
P
is
a
a tangent and the eccentricity of a
conic, the
circle.
be a point on a central conic such that the lines joining
to the foci are at right angles,
CJJ^--2BC-.
Find the position and magnitude of the axes of a hyperbola which has a given line for an asymptote, passes through a given point, and touches a given straight line at a given point. 32.
33.
If
P
and
the incentre of the triangle >SP>S' vertices of the hyperbola. 34.
With two
conjugate diameters of an ellipse as asymptotes
a pair of conjugate hyperbolas are constructed prove that if one hyperbola touch the ellipse the other will do likewise and that the diameters drawn through the points of contact are conjugate to ;
each other. 35.
Prove that a
circle
can be described to touch the four
straight lines joining the foci of a hyperbola to
any two points on
the same branch of the curve.
Tangents are drawn to a hyperbola and the portion of each 36. tangent intercepted by the asymptotes is divided in a given ratio shew that the locus of the point of section is a hyperbola.
;
From a point A' on an asymptote of a hyperbola PB is 37. drawn touching the hyperbola in P, and P2\ PVuve drawn through P parallel to the asymptotes, cutting a diameter in T and V RV is shew that TP and Tj) touch joined, cutting the hyperbola in P, }) ;
;
the hyperbola. [Project the hyperbola into a circle and
V into
the centre.]
CD are conjugate semi-diameters of a hyperbola, and the tangent at P meets an asymptote in L; prove that if PD meet the transverse axis in F, LFC is a right angle. 38.
CP
and
THE HYPERBOLA
191
From a given point on a hyperbola draw a straight line 39. such that the segment between the other intersection with the hyperbola and a given asymptote shall be equal to a given line.
When 40.
does the problem become impossible? If
P and Q
be two points on two circles S^ and *S'o belonging which L is one of the limiting points, such
to a coaxal system of
that the angle
PLQ
is
a right angle, prove that the foot of the
PQ lies on one of the circles of the system, and thus shew that the envelope of PQ is a conic having a focus perpendicular from
L
on
at L. 41.
If a conic
touch the sides of a triangle at the feet of the
perpendiculars from the vertices on the opposite sides, the centre of the conic must be at the
symmedian point
of the triangle.
192
CHAPTER XIV THE RECTANGULAR HYPERBOLA 190. is
A
rectangular hyperbola as
one which has
its
we have already explained
asymptotes at right angles and
its
trans-
and conjugate axes equal. The eccentricity of a rectangular hyperbola = ^2, for e = CS GA, and CS' = CA' + CB' = WA\ verse
:
We
will
now
set forth a series of propositions giving the
chief properties of the curve.
191.
Prop.
1)1
are equal, and if
a rectangular hyperbola conjugate diameter's
QV
he
an ordinate of a diameter POP',
QV"-=PV.P'V.' CP' - CD' = GA ' - GB'
For we have
(§ 1 80)
= and
QV-':PV.PV=GD':GP' = 1.
.
192.
Prop.
(§174)
Gonjugate diameters of a rectangular hyper-
bola are equally inclined to each of the asymptotes.
For the asymptotes are the double lines of the involution by the pairs of conjugate lines through G, and therefore the asymptotes are harmonically conjugate with any Hence as the asymptotes are at pair of conjugate diameters. right angles they must be the bisectors of the angles between pencil formed
each pair of conjugate diameters
Cor.
1.
tangents at asymptotes.
Any its
(§ 72).
diameter of a rectangular hyperbola and the
extremities are equally inclined to each of the
THE RECTANGULAR HYPERBOLA Cor.
Any
2.
chord of a rectangular hyperbola and the
diameter bisecting
193. to the
it
are equally inclined to each asymptote.
Prop. Any diameter of a rectangular hyperbola
diameter perpendicular
This bola
is
193
is
obvious
when we
to it
is
equal
of the conjugate hyperbola.
consider that the conjugate hyper-
in our special case equal to the original hyperbola
and
can be obtained by rotating the whole figure of the hyperbola
through a right angle about an axis through dicular to
its
centre perpen-
plane.
If a hyperbola have two perpendicular diameters equal to the hyperbola itself and the
194. to
its
one another, the one belonging
other to
Let
its
conjugate, the hyperbola
CP
and
CQ
must be a rectangular
one.
be the semi-diameters at right angles to
one another and equal, conjugate.
Q
P
being on the hyperbola, and
Q on
the
THE RECTANGULAR HYPERBOLA
194
PN^':CN'-CA' = BG-':AG'
Now and
(§167)
-BC^ = AC^: BC\ ON"FN'
QM^- CM":
whence we get & and
vtv> ^C-
—BC =
GAP
QM'-
BG^'
AG-
tvtt:
PN''
^^'U^-fil?^, since GN-'
Prop.
195.
Let
it
J.C'
AG-
= J5a
A
the
passes also through the ortliocentre.
P
its
orthocentre and
AD
the
on BG.
Let the rectangular hyperbola meet
AD
again in p.
Ap and BG are at right angles the diameters
Since the chords
them
1
BG-'
+ PN' + 0,
be the triangle,
perpendicular from
parallel to
J
// a i-ectangular hyperbola pass through
of a triangle
ABG
1.
CN-'=GM" and PN''=QM-.
Subtract and use
vertices
=
,
1
will
meet one the hyperbola and the other the
conjugate hyperbola.
Thus and the
DB.DG and
Dp. DA
will
have opposite signs
ratio of their numerical values will
(§ 177),
be unity since the
diameters parallel to them being at right angles are equal. .-.
BD.DG=Dp.DA.
THE RECTANGULAR HYPERBOLA .
.
= .-.
that
where Q is the point in which AD produced meets the circumcircle
BD DC = AD DQ
But
-AD.DP
Dp.
DA = DA
.:
Dp==DP
195
(§6).
.DP.
coincides with P.
is ])
When
Cor.
a rectangular hyperbola circumscribes a
the orthocentre will
lie
vertices lie on one branch
centre will
lie
and the third on the
other, the ortho-
on that branch on which are the two vertices.
lie
Prop.
196.
tri-
on the same branch of the curve, on the other branch, but if two of the
angle, if the three vertices
//'
through the orthocentre
a conic circumscribing a triangle pass it must be a rectangular hyperbola.
Let ABC be the triangle and AD, BE, C'i^the perpendiculai-s
meeting It
in the orthocentre P.
is
clear that the conic
must be a hyperbola,
since
it
is
impossible for two chords of an ellipse or parabola to intersect at a point external to one of
chords
AP
Now
BC= the
BG do
and
since
them and not
to the other,
and the
so intersect.
BD DC = AD .
PD,
.
the diameter parallel to
diameter parallel to AP.
And' these diameters must belong the one to the hyperbola and the other to its conjugate since DB DC and DP DA have .
opposite sign
Therefore the hyperbola
Prop.
197.
.
(§ 177). is
a rectangular one
If a rectangular hyperbola circumscribe
triangle, its centre lies on the nine points circle
Let
ABC be
(§ 194).
the triangle, and D, E,
F
a
of the triangle.
the middle points of
the sides.
Let
be the centre of the rectangular hyperbola and DLL'
an asymptote cutting Since
OF
angles with
AB and AC in
bisects the chord
OIL'
L and
L'.
AB, OF and
AB
make
equal
(§ 192, Cor. 2). .-.
zFOL = zFLO. 13—2
THE RECTANGULAR HYPERBOLA
19()
z EOL' = z EL'O.
Similarly .'.
.'.
zFOE = zALL' + zAL'L = zBAO = ^FI)E.
lies
on the circle round
DEF, which
circle is
the nine
points circle of the triangle. 198.
Prop.
The angle between any chord
angular hyperbola and the tangent at subtended by
on
PQ
P
is
PQ
equal
to
of a rectthe angle
at P', the other end of the diameter through P.
Let the chord PQ and the tangent at P meet the asymptote in R and L. Let V be the middle point of PQ.
THE RECTANGULAR HYPERBOLA
Then
VRC = Z VCR PLC = Z PCL
Z Z
and
(§
2).
(§192, Cor. 1 ).
zLPR = ^CLP-zCRV
.-.
192, Cor.
197
.
= zP6'Z-zFCi^ = zF6'P ==ZQP'P Prop.
199.
(since
CV is
parallel to QP').
Art/ chord of a rectangular hyperbola subany diameter angles ivJiich are equal or
tends at the ends of
supplementary.
Let
QR
be a chord, and
Let the tangents at
and
L',
In cuts
P
PCP' and
a diameter.
F meet
the asymptotes in L,
I
I'.
fig. 1,
QR
where Q and
R
lie
on the same branch and PP'
internally,
zQPL = zQP'P
(§198)
Z RPl = Z RP'P.
and .-.
Z
QPR = supplement =
Fiff.
In cuts
fig. 2,
QR
of
sum
of
and RPl
Fig. 2.
1.
where
QPL
supplement of Z QP'R.
Q and
P
lie
on the same branch and PP'
externally,
ZLPR = ZRFP and .-.
^LPQ = ZQP'P. ^RPQ = zQFP-zRFP = zQFR.
IDS
rectangular hyperbola
thp:
In
fig. 3,
QR
cuts
where Q and
z
In
fig. 4,
QR
cuts
on opposite branches and PP'
QPR = z QPL
+ z LPP' + z P'PR = z QP'P + z PP7' + z PP7' = zQP'R.
where Q and
P
lie
on opposite branches and PP'
QPP = z QPP + z PPP = z QP'P + z LPR
Z QP'P .-.
lie
externally,
z and
R
internally,
=Z
QP'X'
+ z P7^'P = z QP'P' + z RPP'.
zQPR + zQP'R' = z.L'P'P + zLPP' =
2 right
Z
s.
EXERCISES 1.
The portion
tercepted between of contact 2.
If
3.
at
any tangent
asymptotes
is
to a rectangular hyperbola in-
double the distance of
its
point
from the centre.
PJV be the ordinate
hyperbola, and
from
of
its
PG
of
any point
the normal at P, prove
P
on a rectangular
GJ^= NG, and
the tangent
N to the auxiliary circle = PN. If
CK be
any point
are similar.
P
the perpendicular from the centre on the tangent
of a rectangular hyperbola the triangles
PC A, CAK
THE RECTANGULAR HYPERBOLA
PQR
4.
is
P
the angle at
If
a triangle inscribed in a rectangular hyperbola, and a right angle; prove that the tangent at
is
PP' and QQ' be perpendicular chords
hyperbola then PQ', QP' will be at
PP'
6.
any chord
is
perpendicular to
PQP'
P
is
per-
QR.
j^endicular to 5.
199
of a rectangular hyperbola,
meets the hyperbola in
it
of a rectangular
i-ight angles, as also
Q
;
PQ and
P'Q'.
and a diameter
prove that the circle
touches the hyperbola at Q.
from the extremities of any diameter of a rectangular
If
7.
hyperbola lines be drawn to any point on the curve, they wull be equally inclined to each asymptote.
Focal chords of a rectangular hyperbola which are at right
8.
angles to one another are equal.
[See§§ 116, 117.]
The distance
9.
the centre
of
any point on a rectangular hyperbola from mean between its distances from tlie
the geometric
is
foci.
If
10.
PP' be a double ordinate
rectangular hyperbola whose centre
is
to the transverse axis of a
C, then
C P'
is
perpendicular
to the tangent at P.
The centre
11.
of the inscribed circle of a triangle lies
on any
rectangular hyperbola circumscribing the triangle whose vertices are the
centres.
e
Focal chords parallel to conjugate diameters of a rectangular
12.
hyperbola are equal. If the
13.
tangent at any point
P
of a rectangular hyperbola,
centre C, meet a pair of conjugate diameters in the circle 14.
E and F, PC touches
CEF.
Two
hyperbola.
tangents are drawn to the same branch of a rectangular tliat the angles which these tangents subtend at
Prove
the centre are respectively equal to the angles which they
make
with the chord of contact. 15.
A
circle
and a rectangular hyperbola
intersect in
four
common chords is a diameter of the hyperother common chord is a diameter of the circle.
points and one of their
bola
;
shew that the
16.
their foci
Ellipses are described in a given parallelogram lie
on a rectangular hyperbola.
;
shew that
THE RECTANGULAR HYPERBOLA
200 17.
If
hyperbola
from any point Q
QA
The
QR
parallel to the trans-
QR = AQ.
verse to meet the curve, 18.
in the conjugate axis of a rectangular
be drawn to the vertex, and
lines joining the extremities of conjugate diameters of
a rectangular hyperbola are perpendicular to the asymptotes. 19.
The base
of a triangle
being given the locus of 20.
The
its
and the
vertex
circles described
is
difference of its base angles
a rectangular hyperbola.
on parallel chords
of a rectangular
hyperbola are coaxal. 21.
If
a rectangular hyperbola circumscribe a triangle, the
pedal triangle 22.
is
a self-conjugate one.
At any
point
P
of a rectangular hyperbola the radius of
curvature varies as C/*", and the diameter of the curve
is
equal to
the central chord of curvature. 23. is
At
an}' point of a rectangular
24.
FN is drawn perpendicular to an asymptote of a rectangular P on the chord of curvature along PN
hyperbola from any point IS
hyperbola the normal chord
equal to the diameter of curvature.
equal to
-p^
-
it,
201
CHAPTER XV ORTHOGONAL PROJECTION When
200. figure
is
the vertex of projection by
p on to
projected from one plane
means of which a
another plane
vr is
at a
very great distance from these planes, the lines joining corre-
sponding points in the original figure and
What we may
to being parallel.
its
projection
come near
call cylindrical projection is
the
case in which points on the p plane are projected on to the plane by lines which are all drawn parallel to each other.
tt
We
regard this as the limiting case of conical projection
V is
vertex
when the
at infinity.
In the particular case where the lines joining corresponding points are perpendicular to the
on the is
plane
p
tt
plane on to which the figure
projected, the resulting figure on the
is
tt
plane
said to be the orthogonal projection of the original figure.
Points in space which are not necessarily in a plane can be orthogonally projected on to a plane by drawing perpendiculars
from them to the plane.
The
foot of each perpendicular is the
projection of the point from which
which
it is
drawn.
Thus
all
points
on the same line perpendicular to the plane on to which the projection is made will have the same projection. in space
lie
In the present chapter
it
will
be shewn how certain pro-
perties of the ellipse can be obtained from those of the circle, for,
a
as
we
circle.
shall see, every ellipse is the orthogonal projection of It is first necessary to establish certain properties of
orthogonal projection. 201. projection
It
may be
observed at the outset that in orthogonal
we have no vanishing
line as in conical projection.
(ORTHOGONAL PROJECTION
202
The
line at infinity in the
infinity in the plane
ir.
perpendiculars to the
ir
infinity
ji
This
plane projects into the line at is
clear from the fact that the
plane from points in
meet the p plane
it
on the
line at
at infinity.
It follows that the orthogonal projection of a parabola will
be another parabola, and of a hyperbola another hyperbola, while the orthogonal projection of an ellipse
Avill
be another
ellipse or in particular cases a circle.
The
202.
following propositions relating to orthogonal pro-
jection are important
Prop.
Tlie projection
of a straight line
is
another straight
line.
This
is
obvious from the fact that orthogonal
limiting case of conical projection.
is
only a
It is clear that the line in
TT plane which will be the projection of a line I will be that which the plane through I and perjjendicular to the plane tt
the in
cuts this 203.
TT
plane.
Prop. Parallel straight lines p7'oject into 2Ja'>'allel and in the same ratio as regards their length.
straigJit lines,
Let
J^i)
and CI) be
tAvo lines in space parallel to
one another.
ORTHOGONAL PROJECTION Then ab and cd must be a point
p,
p Would
parallel, for if
203
they were to meet in
be the projection of a point
common
to
AB
and CD.
Now draw AF and CG parallel respectively to ab and cd to meet Bb and Dd in F and G. Then AabF is a parallelogram so that AF= ab, and similarly CG = cd. Now
since
AB
is
parallel to
CD, and ^i^
are respectively parallel to ab and cd which
be
parallel),
F and G are
the angle
.'.
A ^i^5
.-.
AF CG = AB
.-.
ab:cd
Prop.
to p's intersection
OH
IT will
Let
be
a
is
similar to
:
Lengths same ratio.
204.
angle
GCD
CG
(for these
;
to
and the angles at
right angles.
Cor. in the
FAB = the
to
we have proved
:
A CG^D.
CD.
= AB:CD.
of line lying along the
//
I
be
with the
a limited ir
same
line in the
line are projected
p plane
parallel
plane, the orthogonal projectdon of
line parallel to
and of
AB be the limited line
I,
the
and ab
same
its
length as
I
I.
orthogonal projection.
204
ORTHO(;OJN'AL PROJECTION
BD
Draw J.Cand of
p and Then
perpendicular to the line of intersection
TT.
ACDB is
Also since
Aa
perpendicular to
a parallelogram.
and Bb are perpendicular to CD and therefore they are
Ga and Dh
tt,
are
parallel to each
other.
AACa = ABDb AC^BD, zAaC = Z BbD /.AGa= ^ BDb for AO and Ca are parallel
Further for
and
.-.
.".
as C'a
and Db are .-.
205.
A
Prop.
perpendicular
bear
BD
and Db.
parallel, (7i)&a is a parallelogram.
ab
= CD = AB.
limited line in the
the line of intersection of
to
Ca = Db.
p and
ir
p
'
plane perpendicular
to
will project into a line also
of intersection and whose length will a ratio equal to the cosine of the angle
to this line
to the original line
between the planes.
Let
AB be
let its line
perpendicular to the intersection of p and
meet
it
tt,
and
in C.
Let ab be the orthogonal projection of AB.
Then ab and
AB
meet
in C,
and ab
:
AB = ac AC :
= cos aCA = cos ( Z between p and
tt).
ORTHOGONAL PROJECTION 206.
A
Prop.
closed figure on the
a closed figure luhose area will bear
a
p
to that
plane
205 ivill
project into
of the original figure
ratio equal to the cosine of the angle betiveen the planes.
For we
number
of"
may
suppose the figure to be made up of an infinite
narrow rectangular
strips the length of
p and
which runs
The lengths
parallel
to the intersection of
slips are
unaltered by projections, and the breadths are diminished
in the ratio of the cosine of the angle
The
207.
We ellipse
and
Now
viz.,
its
(§
of the
between the planes.
ellipse as the orthogonal projection
have seen
another,
tt.
of a
circle.
150) that corresponding ordinates of an
auxiliary circle bear a constant ratio to one
BC:AC.
its major axis comes into a plane making with that of the ellipse an angle whose cosine is BG:AC.
AA'
let
until it
the auxiliary circle be turned about
ORTHOGONAL PROJECTION
206
It is clear that the lines joining each point on the ellipse to
new
the
position of the point corresponding to
circle will
its
it
be perpendicular to the plane of the
Thus the ellipse new position.
on the auxiliary
ellipse.
the orthogonal projection of the circle in
is
Certain properties of the ellipse then can be deduced from those of the circle by orthogonal projection.
some
We
proceed to
illustrations.
Prop.
208.
diameters of an ordinates P'M,
CP
If
and CD he a pair of conjugate semiand CP', CD' another such pair, and the drawn to CP, tlien
ellipse,
D'N he P'M CN = D'N: CAl = CD CP. :
For
let
adjusted to
:
the corresponding points
make an angle
be denoted by small
Then Cp and Cd
in
the auxiliary circle
BC
cos~^ -r--, with the plane of the ellipse,
letters.
are perpendicular radii as are also
and p'm, d'n being Cp, and we have
parallel to
Cd',
Cd
will
A Cmp =Ad 'nC. • .
and
.
p'm Cd = Cn Cp
by
§
:
:
d'n
:
Cd = Cm
:
Cp.
203
P'M:GD=CX:CP D'N:CD=CM:CP.
PM
:
CN - CD CP = D'N :
:
Cp and
be perpendicular to
CM.
ORTHOGONAL PROJECTION
207
Prop. If the tangent at a point P of an 209. any pair of conjugate diameters in T and T' and CD to CP, then rP PT' = CD\
ellipse
meet
he conjugate
.
For in the corresponding figure of the circle Ct and at right angles, and Cp is perpendicular to tt'.
Ct' are
t'
J'
= Cp- = Cd tp:Cd = Cd:pt. TP:CD = CD:PT'. TP.PT' = CD\
.'.
'".
tp pt' .
.-.
Prop.
210.
CA
and
CB
For the
is IT.
make an angle
Area of ellipse
•.
:
on
to
Area of
= BC A C (§ 206 = tt BC .AC. :
).
.
The orthogonal projection of a circle from a tt is an ellipse whose major axis is intersection of p and tt, and equal to the diameter
circle.
Let xiA' be that diameter of the the
circle
ellipse
ellipse.
another plane
jKirallel to the
of the
BC
Prop.
211.
p
semi-axes are
cos'^jy; with that of the
Area of auxiliary .-.
plane
ellipse ichose
ellipse is the orthogonal projection of its auxiliary
circle tilted to
.
The area of an
CA.cn.
p and Let
tt
A A'
circle
which
planes.
project into
aa equal
to
it (§ 20-1).
is
parallel to
ORTHOGONAL PROJECTION
208 Let
PN be
an ordinate to the diameter
A A'
and
let
pn be
its projection.
p/i
.•.
is
= PN cos a
where a
perpendicuhxr to cm'
Now p7i-
:
CD)
na
.
is
the Z between
p and
= PN" cos^ a ^iV NA' :
= cos- a
:
.
1.
Hence the axis,
and
The
its
locus of p is an ellipse having aa minor axis = aa x cos a.
eccentricity
CoR.
1.
into similar
is
easily seen to
major
a.
will
be equal and the major axis of
major axis of the other, each being
parallel to the
is
be sin
for its
Two circles in the same plane project orthogonally and similarly situated ellipses.
For their eccentricities the one
and pn
ir,
(§ 205),
parallel to the line of intersection of the planes.
Cor.
Two
2.
similar and similarly situated ellipses are the
simultaneous orthogonal projections of two
circles.
EXERCISES 1.
The
locus of the middle points of chords of
pass through a fixed point 2.
is
ellipse
which ellipse.
an ellipse its sides are and the greatest area of such a
If a parallelogram be inscribed in
parallel to conjugate diameters,
parallelogram 3.
an
a similar and similarly situated
If
jugate to
is
BC.A<'.
PQ be any CP in 7\
chord of an ellipse meeting the diameter conthen
PQ PT --=2CR.
where
CR
is
the semi-
diameter parallel to PQ. If a variable chord of an ellipse bear a constant ratio to the 4. diameter parallel to it, it will touch anoth(,M- similar ellipse having its axes along those of the original ellipse. 5.
The greatest
has one of
its sides
triangle which can be inscribed in an ellipse
bisected by a diameter of the ellipse
and the
others cut in points of trisection by the conjugate diameter.
ORTHOGONAL PROJECTION 6.
209
meet two concentric, similar and similarly the portions intercepted between the curves are
If a straight line
situated ellipses, equal. 7.
The
locus of the points of intersection of the tangents at the
extremities of pairs of conjugate diameter
and similarly situated 8.
BP,
If
BD
CP,
CD
is
a concentric, similar,
ellipse.
be conjugate serai-diameters of an
be joined, and
AD, A'P
ellipse,
intersect in 0, the figure
and
BDOP
will be a parallelogram.
Two
whose axes are at right angles to one another Shew that any pair of common chords make equal angles with an axis.
9.
ellipses
intersect in four points. will
10.
Shew that a
ellipse itself its circles of
circle
of
curvature for an ellipse and the
can be projected orthogonally into an
ellipse
and one
curvature.
14
of
210
CHAPTER XVI CROSS-RATIO PROPERTIES OF CONICS 212.
and
P
Prop.
If A, B,
a variable point on
G,
D
he
the conic,
four fixed points on a
P (ABCD)
is
conic,
constant
and
corresponding cross-ratio of the four points in which the tangents at A, B, 0, meet that at P. equal
to the
D
Project the conic into a circle and use corresponding small letters in the projection.
P{ABGD)=p{ahcd).
Then But p (abed)
is
constant since the angles apb, bpc, cpd are
constant or change to their supplements as therefore
P {ABCD)
is
Let the tangents at let
p moves on
the circle;
constant. a, b, c,
be the centre of the
d cut that in p in
circle.
a^, b^, Ci, d^
and
CROSS- RATIO PROPERTIES OF CONICS
Then
Ob^, Oc^, Od^ are perpendicular to pa, ph, pc, pd.
Oa-^,
p {abed) =
.•.
.-.
Cor.
211
(ciibiCidi)
=
(ajbiCidi).
P{ABCD) = {A^B,C,D,).
If J.' be a point on the conic near to A,
we have
A'{ABGD) = P{ABGD). .-.
T be
if J.
the tangent at A,
A (TBCD) = P{ABCD). Note. In the special case where the pencil formed by joining any point P on the conic to the four fixed points A, B, C, D is harmonic, we speak of the points on the conic as harmonic. Thus if P {ABCD) = -\, we say that A and C are harmonic conjugates to B and D. Prop.
213.
points in a plane the locus
Let
of
Q
If A, B, C, D be four fixed non-collinear and P a point such that P (ABCD) is constant,
P is a
conic.
be a point such that
Q {ABCD) = P (ABCD).
%
Then
if
the conic through the points A, B, C, D,
pass through Q, let
it
cut
QA
P
does not
in Q'.
.-.
P {ABGD) = Q' (ABCD)
.-.
Q' (ABCD) = Q (ABCD).
by
Thus the pencils Q (A, B, C, D) and homographic and have a common ray QQ'.
§ 212.
Q
(A, B, G,
D)
14—2
are
CROSS- RATIO PROPERTIES OF CONICS
212 Therefore
A, B,
C,
But
(§
D are
64) they are coaxally in perspective
that
is,
this is contrary to hypothesis.
P
Therefore the conic through A, B, C, D,
Thus our proposition
We
;
collinear.
is
see from the above that
the five points A,B, G,D,
goes through Q.
proved.
E as
we may regard
a conic through
P
such that
line wJiich cuts
four non-
the locus of a point
P {ABCD) = E {ABGD). 214.
Prop.
TJie envelope
of a
concurrent coplanar fixed straight lines in four points forming a
range of constant cross-ratio
is
a conic touching
the
four
lines.
This proposition will be seen, when we come to the next chapter, to follow
by Reciprocation
directly from the proposition
of the last paragraph.
The
following
Let the
line
p
is
an independent proof
cut the four non-concurrent lines
the points A, B, G,
D
such that
{ABGD) =
a, b,
c,
d in
the given constant.
213
CROSS- RATIO PROPERTIES OF CONICS
Let the
Then
if
b, c,
lines in A', B',
q be not a tangent to the conic touching
d cut .-.
q'
d,p,
in B", C", D".
{A'B"C"D")
= {ABCD) by = {A' BCD').
The ranges A'B'C'D" and A'B'C'D'
common b, c,
§
212
are therefore
homo-
corresponding point.
Therefore they are in perspective to our hypothesis that a,
(§ 60),
which
is
contrary
d are non-concurrent.
Thus q touches the same conic c,
a, h, c,
a tangent to the conic.
q'
graphic and they have a
a, b,
C, D' such
= (A BCD).
from A' in q draw
Let
same four
line q cut the
that {A'B'C'D')
as
that
which touches
d, p.
And 215.
our proposition
Prop.
a conic S, and A,,
//'
F
is
established.
(A, B, C,
B„ C„
D)
be a pencil in the
plane of
FA, FB, FC,
FD with
D, the poles of
respect to S, then
P{ABCD) = (A,B,CM.
We as
need only prove
we have
this in the case of a circle, into
seen, a conic can be projected.
which
214
CROSS-RATIO PROPERTIES OF CONICS
Let
be the centre of the
Then OA^, OB^, 00^, OD^
are perpendicular respectively to
PR
PA, PB, PC, ..
circle.
P (ABCD) = (A,B,CM = (A.BAD,).
This proposition
is
of the greatest importance for the pur-
poses of Reciprocation.
We
had already seen that the polars of a range of points we now see that the pencil is homographic with
form a pencil
;
the range. 216.
PascaPs theorem.
If a
conic pass through six
points A, B,0, D, E, F, the opposite pairs of sides of each of the sixty different hexagons (st^-sided figures) that can be formed
with these points intersect in collinear points.
This theorem Chap. X). Or we
may may
be proved by projection
(see Ex. 25,
proceed thus
Consider the hexagon or six-sided figure formed with the sides
AB, BG, CD, DE, EF, FA.
CROSS-RATIO PROPERTIES OF CONICS
The pairs
of sides which are called opposite are
BG and EF; CD CD
Then
meet
EF in
Z respectively.
H, and
DE meet FA
A (BDEF) = C
since
.-.
is,
XY,DH and
and
FG= are concurrent
Z, the intersection of
Thus the proposition
The student should different
XDEG
YHEF
have a
corresponding point ^. .-.
that
in G.
(BDEF),
(XDEG) = (YHEF).
These homographic ranges
common
;
and FA.
Let these meet in X, Y, Let
215
AB and DE
is
DH and
FG,
lies
(§ 60),
on
XF.
proved.
satisfy
himself that there are sixty
hexagons that can be formed with the
six
given
vertices.
217.
Brianchon's theorem.
hexagon the
If a
conic be inscribed in a
lines joining opposite vertices are concurrent.
This can be proved after a similar method to that of
and may be
left
as an exercise to the student.
§
216,
We shall content
ourselves with deriving this theorem from Pascal's by Reciprocation.
To the
principles of this important development of
modem Geometry we
shall
come
in the chapter
immediately
following this. 218.
Prop.
fixed points
Let
is
I'he locus of the centres
of conies through four
a conic.
be the centre of one of the conies passing through
the four points A, B, C, D.
516
CROSS-RATIO PROPERTIES OF CONICS
Let M,,M,, M,, M, be the middle points of AB, BC, CD,
DA
respectively.
Draw
Oil//,
OM.:, OM,', OM,' parallel to
DA
AB, BG, CD,
respectively.
Then
Oifi,
OM,'; OM,, OM,'
;
OM;
OM,,
0M„
OM,' are
pairs of conjugate diameters.
Therefore they form an involution pencil.
{M,M,M,M,) =
.-.
But the right-hand
side
is
{M,'M.^M;m:).
OM,
constant since OM,,
&c. are
in fixed directions.
(MJl^MsM,)
.-.
the locus of
.*.
Cor.
The
1.
is
is
constant.
a conic through M,,
conic on which
lies
M„
M^, M,.
passes through ifg,
M^
the middle points of the other two sides of the quadrangle.
For points
if Oi, 0.^,
through
O3, O4, O5
be five positions of 0, these five M„ M,, M, and also on a conic
on a conic through M^,
lie
ilfi,
M„ M„
M^.
But only one conic can be drawn through Therefore is
M„
J/j,
M..,
M^, M^,
M,
all lie
five points.
on one conic, which
the locus of 0.
Cor.
2.
The
locus of
also passes
through P, Q,
R
the
diagonal points of the quadrangle.
For one of the conies through the four points AB, CD and the centre of this conic is P.
lines
So
is
the pair of
;
for
Q and
R.
Prop. If [AA', BB', CC'\ 219. and if a conic he draivn through to cut C, C, then the chords A A', BB' GC are ,
Let
A A'
and BB' intersect
he
an involution pencil
the rays in
A, A', B,
B',
P
for
concurj-ent.
in P.
Project the conic into a circle with the projection of its centre.
Using small
letters in the projection,
are right angles, being in a semicircle.
we
see that aoa, hoh'
CROSS-RATIO PROPERTIES OF CONICS
Hence they determine an orthogonal .•.
coc is a right angle
/.
CC
that
;
is,
cc
217
involution.
goes through
p.
goes through P.
It will be understood that the points
AA'
,
CC
BB',
when
joined to any other point on the conic give an involution pencil for this follows at
once by the application of
§
212.
A
system ^f points such as these on a conic involution range on the conic.
The point called
tlie j)ole
P
is
called an
where the corresponding chords intersect
is
of the involution.
EXERCISES If {P,
1.
Note on
(see
F), (Q, Q') be pairs of harmonic points on a conic § 212), prove that the tangent at P and PP' are
harmonic conjugates to PQ and PQ'. Hence shew that normal at P, PQ and PQ' make equal angles with PP.
if
PP' be
Pand P' and is normal and PQ' are equally inclined to PP' and cut the conic again in Q and Q'. Prove that PQ and P'Q' are harmonic conjugates to and the tangent at P. 2.
at P.
The The
straight line
PP'
straight lines
cuts a conic at
PQ
PP
3.
Shew
that
if
the pencil formed by joining any point on a
conic to four fixed points on the same be harmonic,
two
sides of the
quadrangle formed by the four fixed points are conjugate to each other with respect to the conic.
CROSS-RATIO PROPERTIES OF CONICS
218 4.
vl/,
Deduce from
of a hyperbola intersects the
and the tangents at the vertices in
PAf,5.
P
at any point
The tangent
asymptotes in il/j and Lr,, prove that
L^
and
= PL^.rL,.
Pascal's theorem that
if
a conic pass through
the vertices of a triangle the tangents at these points meet the opposite sides in collinear points.
AA'BB'CC
[Take a hexagon near to ^, B, C] 6.
in the conic so that A', B'
,
C are
Given three points of a hyperbola and the directions of both
asymptotes, find the point of intersection of the curve with a given straight line 7.
drawn
Through a
parallel to one of the asymptotes.
fixed point
on a conic a
line is
drawn cutting the
conic again in P, and the sides of a given inscribed triangle in
A\ B\ 8.
C".
Shew
A, B, C,
that
D
[PA'B'C)
AD
to one asymptote meets
asymptote meets
CB
in
Any two
points
\n
A",
DL KL is
and
Prove that
//.
The sixty Pascal intersect three by three. 9.
10.
constant.
is
are any four points on a hyperbola
CK parallel
;
parallel to the other parallel to
AB.
lines corresponding to six points
D
and
E are taken on
on a conic
a hyperbola of which
CA and CB through and U respectively meet in Q the tangent at D meets CB in P, and the tangent at PJ meets CA in T. Prove that T, Q, R are
the asymptotes are
GA and CB
D
the parallels to
;
;
collinear, lying 1 1.
The
on a
lines
line parallel to
CA and CB
are tangents to a conic at
D and E are two other points ABinG, AE in H, and BE in K.
and
CD^ GD'^CII. :
DE.
on the conic. Prove that
CK GH :
.
The
line
A and B,
CD
cuts
GK.
12. Through a fixed point A on a conic two fixed straight lines AI, AI' are drawn, iS'and S' are two fixed points and P a variable
PS, PS' meet AI, A I' shew that QQ' passes through a fixed point, point on the conic
13.
If
;
two triangles be
in Q, Q' respectively,
in perspective, the six points of inter-
section of their non-corresponding sides
lie
on a conic, and the axis
of perspective is one of the Pascal lines of the six points.
219
CROSS-RATIO PROPERTIES OF CONICS
through a fixed point P are equally inclined to the tangent at P, the chord QQ' passes through a fixed point.
two chords PQ, PQ'
14.
If
15.
If the lines
of a conic
AB, BC, CD, DA touch a conic at P, Q, R, S shew that conies can be inscribed in the hexagons and BQRDSP.
respectively,
APQCRS
The tangent at P to an ellipse meets the auxiliary circle in ASS' A' is the major axis and SY, S'Y' the perpendiculars from the foci. Prove that the points A, Y, Y', A' subtend at any point on the circle a pencil whose ci-oss-ratio is independent of the 16.
Y and
Y'.
position of P. 17.
li A,
diameter, shew
and
PB
B
he given points on a circle, and
how
shall cut
CD
in
P
CD
be a given
on the circle such that points equidistant from the centre.
to find a point
PA
220
CHAPTEE XVII RECIPROCATION 220.
If
we have a number
_an d take t he_p olars p, q,
F
a__conie
^goin ts
in the plane
F
and Q
rg_spect to
and
r
a,g
is .
of th e
i
.
,
r,
fcc.
in a plan e
thenjbhe^m^^jommg^a^y;!^^^
y e- Ji aye
jtlgeady_§een. the_polai^ with
nterse ction of
th, e_c orrespondinff
J ines p
q.
It will
be convenient
to represent the intersection of the
p and q by the symbol and Q by (PQ).
lines
P
of points P, Q, R,
&c. o^_these points with respect to
(pq),
and the
line joining the points
The p oint P corre sponds with the line p^ in t he sense Ihat PJsjb he pole ofj j^cLthfi^line^ C^^ correspond s with th e point ( pq) in the sense that (PQ) is_th e polar o £Xj2i^),
Thus
if
we have
points and lines,
a
fi
P
gure
cojisisti ng of
t hea^jcorrespondinp' toJ.t^
—Two
consistmg^f_lin£Sjand-^>oints called
relation
in
medium
to
one
another
o ftheir_ Recipro city
Using
§
is
snch
we
an aggregate of h av e a GignreJ^'
fignrea^Xand^^re
Recipr ocal
Ji^reS;_JThe
the_conic^J\_
215 we see that a range of points in jP corresponds homographic with the range? in F'.
to a pencil of lines,
221
RECIPROCATION "Rj^mPflTis nf
221.
t,V.p
ciated in the last paragraph
of a figure consisting
property of a
prinniple.
of correspondence enun-
^ve. are
able frOTnji
p oints and
of
lines
fig ure consi st ing of linas
The o ne property
is
kn ov^^^,property
to
another
infer
nnd points.
called the Reciprocal of the other,
the process of passin g from the one to the other
is
and
known
as
Reciprocation.
We
will
give examples.
We know
222.
A'B'C
now
that
if
the vertices of two triangles
ABC,
be in perspective, the pairs of corresponding sides
{BG) (BV), (GA)
(C'A'),
(AB) (A'B')
intersect in collinear
points X, Y, Z.
Fig. F.
Now
if
we draw the
vertices of the triangle
reciprocal figure, corresponding to the
ABG, we have
a triangle whose vertices will be
three lines
(6c), {ca), {ah).
a, h, c
And
forming
similarly
for A'B'C'.
Corresponding to the concurrency of {AA'), {BB'), (CC) the figure F,
we have the
in
collinearity of (aa), (bb'), (cc) in the
figure F'.
Corresponding to the collinearity of the intersections of {B'C'), {CA) {C'A'), {AB) {A'B') in figure F, we have the
(BC)
concurrency of the lines formed by joining the pairs of points (6c) (6'c'), {ca) {ca), {ah) {a'b') in the figure F'.
222
RECIPROCATION
Thus the theorem
of the figure
F
reciprocates into the
following If
two triangles whose sides are
respectively be
abc, a'h'c
such that the three intersections of the corresponding sides are
then the lines joining corresponding vertices,
collinear,
{ah)
and
{ah'), {he)
and
(6'c'),
(ca)
viz.
and {ca), are concurrent.
(ab)
Fig.F'.
The two stated thus
reciprocal,
theorems placed side by side
may be
:
Triangles in perspective are
I
coaxal.
Coaxal triangles are in perspective.
I
The student
will of course
have realised that a triangle
regarded as three lines does not reciprocate into another triangle regarded as three
and
lines,
but into one regarded as three points
vice versa.
223.
harmonic
Let us now connect together by reciprocation the property of the quadrilateral and that of the
quadrangle.
Let
a, h,
c,
d be the
lines of the quadrilateral
corresponding points of the quadrangle.
;
A, B,
C,
D
the
RECIPROCATION
223
(ac)
(bd)
Let the
line joining (ub)
„
„
„
„
„
„
and (cd) be
and {hd) be (ad) and (he) be {ac)
Fig. F'.
p, q, r.
224
RECIPROCATION
The harmonic property
of the quadrilateral
is
expressed
symbolically thus {{ab){cd), {pr)(pq)]
The
= -l,
{(ad) (be),
(pr){qr)}=-l,
{{ac){hd),
(pq){qr)}=-l.
reciprocation gives
= -1, [{AD){BC\ (PR){QR)} = -1,
{{AB)iCD), (FR){PQ)}
{(AC){BD), iPQ)(QR)] = -l. If these be interpreted on the figure
property of the quadrangle,
viz.
we have the harmonic
that the two
of the
sides
diagonal triangle at each vertex are harmonic conjugates with the two sides of the quadrangle which pass through that vertex.
The student
sees
now
that
the
'
diagonal points
'
of a
quadrangle are the reciprocals of the diagonal lines of the quadrilateral '
from which
it
is
Hence
derived.
the
term
diagonal points.'
Prop.
224. to
a.
conic into
For
let
A71 involution range reciprocates with respect-
an involution
pencil.
the involution range be
A, A,; B,B,; C,G,kc. on a line
The c, Ci
p.
pencil obtained
by reciprocation
will
be
a,
a^
;
h,
h^;
&c. through a point P.
= {ABGA,) = (A,B,C,A) by (ABCA,) = {A,B,C, A) by {ahca^) = (a^biCia).
Also
{ahca,)
and
{aAc.a)
But
§
215.
§ 78.
.".
Thus the 225.
pencil
is
Involution
in involution.
property of the
quadrangle
and
quadrilateral.
Prop. Ally transversal cuts the pairs of opposite sides of a quadrangle in pairs of points which are in involution.
•
225
RECIPROCATION Let
ABCD
be the quadrangle
(§ 76).
A,
Let a transversal
^
cut
the opposite pairs of sides
AB, CD in E, E^, AG, BD in F,F„
AD, BG Let
AD
and
BG meet
in G, G,.
in P.
{GEFG,) = A{GEFG,) = (FBCG,)
Then
= D{PBGG,) = {GF,E,G,) = Gt] £", Fi Or) by (
interchanging the
letters in pairs.
ad
(aB)
Hence E, A. G.
£",
;
F, F^
;
G, Gi belong to the
same
I
involution. 15
RECIPROCATION
226
We
havL' only to reciprocate the
above theorem to obtain
other:
tliis
The lines joining any point to the 2mirs of opposite a complete quadrilateral form a j^encil in involution.
Thus
in our figure T,
which corresponds to the transversal
joined to the opposite pairs of vertices (ab), (cd) gives
an involution
Prop.
226.
(ac), (bd)
;
of
t,
(ad), (be)
;
pencil.
circles described
IVie
vertices
on the three diagonals
of a complete quadrilateral are coaxal.
Let AB, BC, CD,
DA
The diagonals
AC, BD, EF.
Let
BD .-.
P
are
be the four sides of the quadrilateral.
be a point of intersection of the circles on
AG and
as diameters.
APC and BPD are
PD; PE, PF are in involution EPF is a ri^ht angle. on EF as diameter goes through P.
But PA, PC; PB, .-.
by
.•.
the circle
§
86 Z
right angles.
(§
225).
i
I
'
RECIPROCATIOX Similarly the circle on
EF goes through BD and AC.
22/ the other point of
intersection of the circles on
That Cor.
is,
the three circles are coaxal.
The middle
points of the three diagonals of a quadri-
lateral are collinear.
This important and well-known property follows at once, since these middle points are the centres of three coaxal circles.
The
line containing these
middle points
sometimes called
is
the diameter of the quadrilateral. 227.
Desargues' theorem.
Conies through four gicen points are cut by any transversal in iniirs
of points belonging
Let a transversal A, B,
G,D
in
P
t
to the
same
involution.
cut a conic through the four })oints
and P,.
Let the same transversal cut the two pairs of opposite sides BD of the quadianglo in E, E, F, F,.
AB, CD; AC,
We
:
now have
{PEF1\) = A{PEFI\) = A {PBCP,)
= D(PBCP,)hy^2l2 = (PF,EJ\) = (PiEiF^P) by interchanging
the
letters in pairs. .-.
P, Pi belong to the involution determined by E, E^
F, F,.
15—2
;
RECIPROCATION
228 'I'hus all t
ABCD
the conies through
in pairs of points
will cut the transversal
belonging to the same involution.
Note that the proposition if the two
Desargues' theorem,
of § 225 lines
is
only a special case of
AD, BC
be regarded as one
of the conies through the four points.
As we
228.
theorem
shall presently see, the reciprocal of Desargues'
the following
is
If conies touch four given lines the jKiirs of tangents to them from any point in their plane belong to the same involution pencil, namely that determined by the lines joining the point to the pairs of opposite vertices of the quadrilateral formed by the
four
lines.
Reciprocation applied to conies.
We
229.
Reciprocation
are is
now going
(jn
to explain
how the
principle of
applied to conies.
Suppose the point P describes a curve S in the plane of the the line p, which is the polar of F with regard to r> will envelope some curve which we will denote by S'.
C(Hiic r,
Tangents to S' then correspond
S
observe further that tangents to
For
P and
let
to points
P' be two near points on
be the corresponding
on S
;
but we must
correspond to points on S,
and
let
>S".
p and
p'
lines.
at
P
point (pp) corresponds to the line (PP') P' moves up to P, (PP') becomes the tangent to and at the same time {p}^') becomes the point of contact
of
p
with
Then the
Now
as
>S'
its
Hence
envelope.
to tangents of
Each of
the curves
*S'
correspond points on
S and
S'
is
S'.
called the polar reciprocal of
the other with respect to the conic T.
230.
point on
Then
If S
Prop.
Let A, B,
C,
D
be
a conic then S'
is
he four fixed points on
S.
P {ABCD)
is
constant.
another conic. S,
and
P any
other
RECIPROCATION
P (ABOD) =
But
229 by
{(pa) (pb) (pc) (pd)}
.-.
{(pa) (pb) (pc) (pel)]
.•.
the envelope of
is
215.
a conic touching the lines
is
j3
§
constant. a, b, c,
d
(§ 214).
Hence S'
is
a conic.
This important proposition might
have been proved as
follows.
'being a conic,
straight lines in
its
is
a curve of the second order, that
plane cut
it
two and only two points,
in
is,
real
or imaginary.
Therefore S' must be a curve of the second
curve such that from each point in
tangents can be drawn to
;
that
If S and S'
Prop.
231.
it
its
is,
S'
is
to
polar and pole of S' and
F
Let
[It is
and
TUhe
that
a
a conic.
and polar of S
corre-
vice versa.
pole and polar of S.
most important that the student should understand
TU is the polar of P with respect to S, not to F. P with respect to F is the line we denote by p.]
that
is
be two conies reciprocal to each
other tvith respect to a conic F, then pole
spond
class,
plane two and only two
The
polar of
Let
QR
be any chord of
tangents at
Q and
R
meet
S which
passes through
in the line
Therefore in the reciprocal figure that
if
any point
tangents from .'.
p and
it
TU,
at
p and
T
P
:
then the
say.
(tii)
are so related
(qt) be taken on p, the chord of contact
to S' passes through (tu).
(tu) are polar
and pole with respect
to
S\
t
of
RECIPROCATION
230 Cor.
Conjugate
1.
jxiints of
S
reciprocate into conjugate
lines of S' an(i vice versa.
Cor.
A
2.
We
232. lei 1.
will
now
some
set forth
If a conic (i.e.
reciprocal theorems in
the conic
contact of
with the
opposite sides are concurrent.
(acj
a conic be circnmscribed
If
be inscribed in
a three-side figure),
the joining lines of the vertices of the triangle and the points of
to
a
triangle
sides
three-point
(a
tho intersections of the
figure),
the triangle with the
of
tangents at the opposite vertices are collinear.
(ab)
six points of inter-
The
section with the sides of a triangle
vertices
of the lines joining the opposite
points
The
two
vertices to
fixed points lie
joining
the
tx'iangle
to
the
intersection
of
the
six
lines
a
of of
opposite sides and two fixed lines
envelope a conic.
on a conic. 3.
will reciprocate into
S'.
columns.
a triangle
2.
S
self-conjugate triangle of
a self-conjugate triangle of
The three points
The three
of inter-
lines joining
the
section of the opposite sides of
opposite vertices of each of the
each of the six-side figures formed by joining six points on a conic
six-point figures
are collinear.
PascaPs theorem.
intersections
a
conic
of
formed by the lines
are
touching
concurrent.
—
Brianchon's theorern. 4.
If
a conic circumscribe a
quadrangle, the triangle formed
by
its
diagonal
points
conjugate for the conic.
is
self-
If a conic
be inscribed in a
quadrilateral, the triangle formed
by
its
diagonals
for the conic.
is
self-conjugate
RECIPROCATION
Prop.
233.
The conic S'
bola, according as the centre
an
is
F
of
231
ellipse,
parabola or hyperor tvithout S.
is within, on,
For the centre of F reciprocates into the line at infinity, and lines through the centre of F into points on the line at infinity.
Hence tangents
to
S
from the centre of
F
will reciprocate
into points at infinity on S',
S
tangents to
Hence
if
will
and the points of contact of these reciprocate into the asymptotes of S'.
the centre of
asymptotes and therefore If the centre of line at infinity, that
If the centre of is
therefore an
F
be outside S, S' has two real
a hyperbola.
be on S, S' has one asymptote,
is,
F be
viz.
the
a parabola.
within
*S',
S' has no real asymptote and
el
Case where F
234.
F
is
a circle.
is
F be a circle (in which case we by C and its centre by 0) a further relation exists between the two figures F and F' which does not otherwise If the auxiliary ov base conic
denote
will
it
obtain.
to
The polar of a point P with respect to G being perpendicular OP, we see that all the lines of the figure F or F' are to the corresponding points
perpendicular to the lines joining of the figure F' or F.
And is
thus the angle between any two lines in the one figure
equal to the angle subtended at
by the
line joining the
corresponding points in the other.
In particular to
S
it
may be
noticed that
are at right angles, then S'
For
if
OP
and
OQ
is
if
the tangents from
a rectangular hyperbola.
are the tangents to S, the asymptotes of
S' are the polars of
P and Q
at right angles since
POQ
is
with respect to C, and these are
a right angle.
If then a parabola be reciprocated with respect to a circle whose centre is on the directrix, or a central conic be reciprocated with respect to a circle with its centre on the director circle,
a rectangular hyperbola
Further
let it
is
always obtained.
be noticed that a triangle whose orthocentre
RECIPROCATION
232 is
at
reciprocate into another triangle also having its
will
orthocentre
at
This
0.
the
student can
easily
verify
for
himself. It can
235.
are connected
that the two following propositions
The orthocentre of a triangle circumscribing a parabola
1. 'lies
now be seen
by reciprocation
on the directrix.
The orthocentre of a triangle inscribed in a rectangular
2.
hyperbola
lies
on the curve.
These two propositions have been proved independently
(i
95, 130).
by
reciprocation.
Let us now see how the second can be derived from the
first
Let the truth of (1) be assumed. Reciprocate with respect to a circle
G having
its
centre
at the orthocentre of the triangle.
Now
the parabola touches the line at infinity, therefore the
pole of the line at infinity with respect to C,
viz. 0, lies
on the
reciprocal curve.
And is
the reciprocal curve
is
a rectangular hyperbola because
on the directrix of the parabola.
Further
is
also the orthocentre of the reciprocal of the
triangle circumscribing the parabola.
Thus we
see that
if
a rectangular hyperbola be circum-
scribed to a triangle, the orthocentre lies on the curve. It is also clear that no conies but rectangnlar hyperbolas
through the vertices of a triangle and
Prop.
236. respect to
a
circle
If 8
G
Let
p)
and we
reciprocate luith
a conic having
S.
be any tangent to S,
P
circle
luhose centre is 0, S' ivill be
for a focus. Let A be the centre of Let
a
be
can
its orthocentre.
Q
its
point of contact.
be the pole of p, and a the polar of
to G.
Draw P3I perpendicular
to a.
A
with respect
RECIPROCATIOX
Then
since
AQ
233
perpendicular to jh "e have by Salmon's
is
theorem (§17)
qp_PM 0A~ AQ•
Thus the is
•
TTiT= the FM
locus of
P
a conic whose focus
constant ^rr,
AQ
which is
•
a point on the reciprocal curve and corresponding directrix the
is
0,
polar of the centre of S.
Since the eccentricity of S' ellipse, parabola, or
without
This
>S'.
is
OA
that
AQ
hyperbola according as in
agreement with
§
is
>S'
is
an
within, on, or
233.
Cor. The polar reciprocal of a coilic with respect to a circle having its centre at a focus of the conic is a circle, whose centre is
the reciprocal of the corresponding directrix.
Let us now reciprocate with respect to a circle the in a semicircle is a right angle.
237.
theorem that the angle Let
A
be the centre of S,
KL
any diameter,
Q any
point on
the circumference.
In the reciprocal figure we have corresponding to directrix a,
and a point
{kl)
on
it
corresponds to {KL).
A
the
RECIPROCATION
234
and I are tangents from {kl) to >S" which correspondcorresponding to Q. and L, and q is tlie tangent to /,;
to
K
*S''
Now {QK)
and \QL) are at right angles.
Therefore the line joining {qk) and
angle at
the focus of
Hence the tangent
reciprocal
subtends a right
theorem
is
that the intercept on anij
conic between tiuo tangents which intersect in the
a
to
{ql)
/S".
directrix subtends a right angle at the focus.
Prop.
238.
A
system,
of non-intersecting coaxal
circles
can
be reciprocated into confocal conies.
Let
L and
L' be the limiting points of the system of circles.
Reciprocate with respect to a circle
Then one
all
C whose
centre
is
the circles will reciprocate into conies having
at L.
L
for
focus.
Moreover the centre of the reciprocal conic of any one of the circles circle.
is
the reciprocal of the polar of
L with
respect to that
RECIPROCATION
But the polar line
of
L
235 same,
for all the circles is the
through L' perpendicular to the line of centres
Therefore
all
the reciprocal conies have a
common
well as a
Therefore they
viz,
the
(§ 22).
common
centre as
focus.
have a second
all
common
focus, that
is,
they
are confocal.
We know that
239.
if
^
be;
common tangent
a
them
of the coaxal system touching
at
P
and
Q,
to
PQ
two
circles
subtends a
right angle at L.
Now
reciprocate this with regard to a circle with its centre
The two
L.
conies, the
circles
of the confocals,
common
P
and
Q
common
point
into the tangents to
point.
confocal conies cut at right angles.
This fact
We
reciprocates into a
t
and the points
the confocals at the
Hence
of the system reciprocate into confocal
common tangent
is
of course
known and
easily
proved otherwise.
are here merely illustrating the principles of reciprocation.
Again
.240.
it is
known
(see Ex.
40 of Chap. XIII) that
if
be two circles, L one of the limiting points, and P and Q points on S^ and ^2 respectively such that PLQ is a right angle, the envelope of PQ is a conic having a focus at L. Si
and
S.2
Now having
reciprocate
its
this
centre at L.
>S\
property with
and
S.,
respect
to
a circle
reciprocate into confocals with
236
L
RECIPROCATION
as one focus
/S/
and SJ,
line
{PQ)
;
the points
p and
viz.
q,
F
and Q
which
reciprocate.' into
tangents to
be at right angles
will
;
and the
reciprocates into the point (pq).
As the envelope
of
{PQ)
is
follows that the locus of {jiq)
a conic with a focus at L,
it
a circle.
is
Hence we have the theorem If
tivo
tangents
from a point
he at right angles, the locus of
This also
is
T, one to each of two confocals,
T is
a
circle.
a well-known property of confocals.
241. We will conclude this chapter by proving two theorems, the one having reference to two triangles which are self-conjugate for a conic, the other to two triangles reciprocal for a conic.
Prop.
//' tivo triangles be self-conjugate to the
same
and
a
their six vertices lie on a conic
Let
ABC, A'B'C be
their six sides touch
conic
conic.
the two triangles self-conjugate with
respect to a conic S. Project
8
into a circle with
then (using small letters
for
A
projected into the centre
the projections) ah, ac are conjugate
diameters and are therefore at right angles, and h and
c lie
on
the line at infinity.
Further .•.
a'h'c' is
a triangle self-conjugate for the
a the centre of the
triangle.
circle
is
circle.
the orthocentre of this
RECIPROCATION Let a conic be placed through the
and
'237
five points a, U, c,
a
h.
This must be a rectangular hyperbola, since as we have seen no conies but rectangular hyperbolas can pass through the vertices of a triangle
.•.
c also lies
and
orthocentre.
its
on the conic through the
five
points
named
above, since the line joining the two points at infinity on a
rectangular hyperbola must subtend a right angle at any point.
Hence the ,•.
six points a,
h, c,
a,
h',
the six points A, B, C, A' B', ,
The second
part
the
of
242.
Prop.
If two
all lie
on a
also lie
proposition
we have
reciprocating this which
c
C
conic.
on a conic.
follows
at
once by
just proved.
triangles are reciprocal for a conic,
they are in perspective.
Let
ABC, A'B'C be two
triangles which are reciprocal for
A
B
the conic S; that is to say, is the pole of B'C, the pole of G'A', C the pole of A'B and consequently also A' is the pole ;
of
BC, B' of CA, and C" of AB.
.•.
Project >S' into a circle with the projection of B' and 6" are projected to infinity.
A
for its centre.
288
RECIPROCATION
Using small is
the pole of
he,
Also since hh'
which
letters for the projection,
aa'
the pole of ac, ah'
h' is
is
perpendicular to
is
we
see that, since a'
he.
is
perpendicular to ac
:
parallel to ah' is perpendicular to ac.
Similarly cc
is
perpendicular to ah.
.'.
aa', hh', cc
meet
.'.
A A',
CC
BB',
in the orthocentre of the triangle ahc.
are concurrent.
EXERCISES 1.
If tlTe conies
*S'
and
the conic V, the centre of of
r with 2.
respect to
aS"
*S"
be reciprocal polars with respect to
corresponds to the polar of the centre
aS'.
Parallel lines reciprocate into points collinear with the centi-e
of the base conic V. 3.
Shew
parallelogram.
that
a
quadrangle
can
be
reciprocated
into
a
239
RECIPROCATION Reciprocate with respect to any conic the theorem
4.
:
The
locus of the poles of a given line with respect to conies passing-
through four fixed points
a conic.
is
Reciprocate with respect to a circle the theorems contained in
Exx. 5
— 12
inclusive.
vertices of a triangle on the
The perpendiculars from the
5.
opposite sides are concurrent.
The tangent
6.
to a circle is perpendicular to the radius thi-ough
the point of contact.
Angles in the same segment of a
7.
The opposite angles
8.
circle are equal.
of a quadrilateral inscribed in a circle
are together equal to two right angles.
The angle between the tangent
9.
a chord through that point
is
at
any point
of a circle
and
equal to the angle in the alternate
.segment of the circle.
The polar
10.
of a point with
respect to a circle
is
perpen-
dicular to the line joining the point to the centre of the circle.
The
11.
locus of the intersection of tangents to a circle which
cut at a given angle
is
Chords of a
12.
a concentric
circle
circle.
which subtend a constant angle at the
centre envelope a concentric circle.
Two
13.
having double contact
conies
will reciprocate
into
conies having double contact.
A
14.
conic
*S".
circle
.S'
is
reciprocated by
Prove that the radius of
the radius of
IS
C
is
means
of a circle
the geometric
and the semi-latus rectum
C
into a
mean between
of 6".
Prove that with a given point as focus four conies can be drawn circumscribing a given triangle, and that the sum of the latera recta of three of them will equal the latus rectum of the 15.
fourth.
Conies have a focus and a pair of tangents
1 6.
common
;
prove
that the corresponding directrices will pass through a fixed point,
and
all
the centres
17.
centre at
lie
on the same straight
line.
Prove, by reciprocating with respect to a circle with *S',
the theorem
:
If a triangle
ABC
its
circumscribe a parabola
KECIPROCATION
240 whose focus to SA, SB,
the lines through A, B,
C perpendicular
respectively
concurrent.
Conies are described with one of their foci at a fixed point two fixed points subtends
18. ^S",
is S,
SC are
so that each of the four tangents from
Prove that the directrices the same angle of given magnitude at S. corresponding to the focus «S' pass through a fixed point. be any point on the common tangent to two parabolas If 19. with a common focus, prove that the angle between the other to the parabolas is equal to the angle between the tangents from axes of the parabolas.
A conic circumscribes the triangle ABC, and has one focus at
20.
dicular to
shew that the corresponding directi-ix is perpen10 and meets it in a point -V such that 10 0X= AO OD,
where /
the centre of the inscribed circle of the triangle, and
0, the orthocentre
;
.
is
A on BC.
the foot of the perpendicular from the centre of the conic.
Prove that chords
21.
of a conic
Shew
.
also
how
D
is
to find
which subtend a constant angle
at a given point on the conic will envelope a conic.
[Reciprocate into a parabola by means of a circle having
its
centre at the fixed point.] If a triangle be i-eciprocated with respect to a circle
22.
on the circumcircle
of the triangle, the point
its
centre
lie
on the circumcircle of the reciprocal triangle.
having
will also
Prove tlie following and obtain from it by reciprocation 23. a theorem applicable to coaxal circles If from any point pairs of tangents p, p q, q, be drawn to two confocals S-^ and S.;,, the angle between p and q is equal to the angle between p and q. :
;
Prove and
24.
following
:
respect to
P, Q,
R
If
anj'^
again in
intersect 26.
respect
to
any conic the
are collinear.
the extremities circle
with
be a triangle, and if the polars oi A, B, C with conic meet the opposite sides in P, Q, R, then
A fixed
25.
reciprocate
ABC
on a
in the plane of a given circle is joined to
point
A and B
P
and
Q.
of
any diameter, and OA,
Shew
OB
that the tangents at
meet the
P and Q
fixed line parallel to the polar of 0.
All conies through four fixed points can be projected into
rectangular hyperbolas.
RECIPROCATION 27.
If
two
of perspective
241
triangles be reciprocal for a conic
is
(§
242) their centre
the pole of the axis of perspective with regard to
the conic. 28. Prove that the envelope of chords of an subtend a right angle at the centre is a concentric
ellipse
which
circle.
[Recipi'ocate with respect to a circle having its centre at the
centre of the ellipse.] 29.
ABC
is
a triangle, /its incentre
;
A^, L\, C\ the points of
Prove that the line joining AA^, BBy, CCi is perpendicular to
contact of the incircle with the sides.
/
to the point of concurrency of
the line of collinearity of the intersections of BiC\,
BC
;
C^A^,
CA
A^By, AB.
[Use Ex.
27.]
16
;
242
CHAPTER CIRCULAR POINTS.
We
243.
XVIII FOCI OF CONICS
have seen that pairs of concurrent lines which are
conjugate for a conic form an involution, of which the tangents
from the point of concurrency are the double
Thus conjugate diameters
the double lines of the involution are
Now
lines.
of a conic are in involution,
and
asymptotes.
its
the conjugate diameters of a circle are orthogonal.
Thus the asymptotes
of a circle are the imaginary double
lines of the orthogonal involution at its centre.
But
clearly the double lines of the orthogonal involution at
one point must be parallel to the double lines of the orthogonal involution at another, seeing that lation, without rotation,
move one
Hence the asymptotes to the
Let
h'
of one circle are, each to each, parallel
asymptotes of any other
C, then
a, h
we may hy a motion of transinto the position of the other.
circle in its plane.
be the asymptotes of one
circle G, a, h' of
a being parallel meet on the line at being parallel meet on the line at infinity. a,
But a and a meet C and h and h' C and „
and
Therefore
on the
Q and
C
another
infinity,
and
h,
C" on the line at infinity,
C
„
„
„
go through the same two imaginary points
line at infinity.
Our
conclusion then
is
that
all circles
the same two imaginary points on the
two points are called the circular points.
in a plane go through
lirte
at infinity.
These
the circular points at infinity or, simply,
CIRCULAR POINTS. .
The
FOCI OF CONICS
243
circular lines at any point are the lines joining that
point to the circular points at infinity
and they are the imaginary double lines of the orthogonal involution at that point.
Analytical point of view.
244. It
point
may if
;
we
help the student to think of the circular lines at any
moment
digress for a
upon the Analytical
to touch
aspect of them.
The equation
of a circle refeired to its centre X-
The asymptotes
is
=
y-
of the form
is
a-.
of this circle are X-
that
-f
+
?/2
= 0,
the pair of imaginary lines
= ix and y = — ix.
y These two
lines are the circular lines at the centre of the
circle.
The
points where they meet the line at infinity are the
circular points. If
we
rotate the axes of coordinates at the centre of a circle
through any angle, keeping them
still
rectangular, the equation
of the circle does not alter in form, so that the asymptotes will
make
angles tan~'(i) and tan~'(—
new
with the
i)
axis of
x as
well as with the old.
This at
first
sight
is
But the paradox
paradoxical.
is
ex-
plained by the fact that the line y = ix makes the same angle tan~^ {%) with every line in the plane.
For
let
y
= mx
Then the angle
be any other that y
positive sense from y
tan~'
—-
;-
mx,
W + imj :;
=
line
through the origin.
= ix makes
with
this,
measured
in the
is
tan
^
{ \
-, 1
-.
—
+tm
- \
=
tan
^
i.
)
Prop. 245. // AOB be an angle of constant magnitude and n, n' he the circular points, the cross-ratios of the jJencil (fl, n', A, B) are constant. 16-2
CIRCULAR POINTS.
244
n
^ (nQ.'AB) ,r^r^> i,s = ^ 4
For
FOCI OF CONICS
—HOCl' „^., nOB
—^AOB
sin
sin -.
sin
but the angles nOfl', flOB, AOi^' are circular lines
and Z
Of i
constant since the
all
make the same angle with every
AOB is
line in the plane,
constant by hypothesis. .-.
Prop.
246.
tvy7v^>
.
sni
0(nn'.4i?)
is
constant.
All cunics passing throuc/h the circular points
(fre circles.
C
Let points,
be the centre of a conic
which we
Then
will
CO., CD.' are the
>S'
O
denote by
passing through the circular Q'.
and
asymptotes of
But the asymptotes are the double formed by pairs of conjugate diameters.
And that
is
>S'.
lines of the involution
the double lines completely determine an involution,
to say there can be only one involuti<)n with the
double
same
lines.
Thus the conjugate diameters
Hence ^
is
of
S
are
all
orthogonal.
.
a circle.
The circular points may he utilised for establishing properties of conies passing through two or more fixed points.
For a system of conies
all
passing through the same twO'
points can be projected into circles simultaneously.
This
is
effected
by projecting the two points into the
points on the plane of projection. will
now go through
they are
The
circular
projections of the conies
new plane and
so
of course understands that such a projection
is
the circular points in the
all circles.
The student an imaginary
one.
We
will
247.
now proceed
to
an illustration of the use of
the circular points. It can
be seen at once that any transversal
is
cut by a system
of coaxal circles in pairs of points in involution (the centre of this involution being the point of intersection of the line with
the axis of the system).
FOCI OF CONICS
CIRCULAR POINTS.
From
this follows at once Desargues'
theorem
245
(§
227),
namely
that conies through four points cut any transversal in pairs of points in involution.
For
if
we
the conies
project two of the points into the circular points
become
all
Moreover the
circles.
coaxal system, for they have two other points in
Hence Desargues' theorem
is
form a
circles
common.
seen to follow from the involu-
tion property of coaxal circles.
The
involution property of coaxal circles again
is
a particular
case of Desargues' theorem, for coaxal circles have four points in
common, two being the
which
all
We
248.
circular points,
and two the points
in
the circles are cut by the axis of the system. will
now make use
of the circular points to prove
the theorem: If a triangle he self-conjugate to a rectangular hyperbola its circumcircle passes through the centre of the hyperbola.
Let
be the centre of the rectangular hyperbola,
self-conjugate triangle, O,
Now
observe
first
that
fl'
ABC the
the circular points.
OHO'
is
a self-conjugate triangle for
For OD., Ofl' are the double lines of the orthogonal involution at to which the asymptotes, being the rectangular hyperbola.
at right
OH, OH' belong
Therefore
angles, belong.
involution whose double lines are the asymptotes
(§ 82),
to the
that
is
the involution formed by pairs of conjugate lines through 0. .•.
which ,•.
OD., is
OnO'
Also .-.
and
are conjugate lines, and
is
is
the pole of ClQ'
a self-conjugate triangle.
ABC is
a self-conjugate triangle.
the six points A, B, C, 0,
must be a
this conic .•.
A, B,
Cor. its
on'
the line at infinity.
centre
C,
ft, D,' all lie
circle as it passes
on a conic
through
ft
(§
241)
and
-
ft'.
are concyclic.
If a rectangular hyperbola circumscribe a triangle, lies
on the nine-points
This well-known theorem proposition, for the pedal
rectangular hyperbola.
is
circle.
a particular case of the above
triangle
is
self-conjugate
(Ex. 21, Chapter XIV.)
for
the
CIRCULAR POINTS.
246
Prop.
249.
Concentric
FOCI OF CONICS circles
have double
contact
at
infinity.
For
be the centre of the
if
points at infinity,
n
and
f)
That is, and n'.
all
circles, D,,
Vl'
the circular
the circles touch Ofl and Ofl' at the points
O'.
the circles touch one another at the points
all
Foci of Conies.
250.
Prop.
Every conic has four foci, tiuo of which lie on one and are real, and two on the other axis and are
axis of the conic
imaginary. Since conjugate lines at a focus form an orthogonal involution, and since the tangents from any point are the double lines of the involution formed by the conjugate lines there, it follows that the circular lines through a focus are the tangents to the conic from that point.
But the
circular lines at
any point go through
il
and H'
the circular points.
Thus the
foci
tangents from
H
by drawing and taking their four
of the conic will be obtained
and iV
to the conic,
points of intersection.
Hence there
are four
foci.
To help the imagination, construct a were
figure as if
H
and H'
real points.
Draw S, 8', F,
tangents from these points to the conic and let F' be their points of intersection as in the figure >S, S' ;
being opposite vertices as also
F and
Let FF' and SS' intersect in
Now is
F'.
0.
the triangle formed by the diagonals FF', SS' and
self-conjugate for the conic, because
the quadrilateral.
(Reciprocal of
.•.
is
the pole of HO',
.•.
is
the centre of the conic.
i.e.
§
119
it
HO'
touches the sides of
a.)
of the line at infinity.
CIRCULAR POINTS. Further OflCl'
is
FOCI OF CONICS
247
the diagonal, or harmonic, triangle of the
quadra^^Ze SS'FF'.
0(nn',FS)==-h
.-.
.'.
OF and OS are
Ofl and on' are the double .'.
OF and OS
(§76)
conjugate lines in the inA^olution of which lines.
are at right angles.
And OF and OS
are conjugate lines for the conic since the
by the diagonals FF', SS', On' is self-conjugate and is, as we have seen, the centre.
triangle formed for the conic .•.
;
OF and
OS, being orthogonal conjugate diameters, are the
axes.
Thus we have two
pairs of
foci,
one on one axis and the
other on the other axis.
Now we know
that two of the
It follows that the other two, if
F were
real,
real point, .•.
which
F and
Cor.
is
FS
say
S and
>S",
are real.
F', are imaginary.
would meet the
For
line at infinity in
a
not the case.
F' must be imaginary.
The
to the conic
coney clic.
the line
foci,
F and
lines joining non-corresponding foci are tangents
and the points of contact of these tangents are
248
CIRCULAR POINTS.
A
Prop.
251.
FOCI OF CONICS
system of conies touching the sides of a
quadrilateral can he projected into confocal conies.
Let
ABCD be the quadrilateral, the pairs of opposite vertices
being A, C;
B,D; E,F.
E and F into
Project
the circular points at infinity on the
plane of projection. .'.
A,
and B, D by § 250.
C
projection,
Cor.
project into the foci of the conies in the
Confocal conies form a system of conies touching four
lines.
252.
We
will
now make
use of the notions of this chapter
to prove the following theorem,
If
which
is
not unimportant.
the sides of two triangles all touch the
vertices
Let
of the triangles
ABC, A'B'C
touch the same conic
all lie on
a
same
conic, the six
conic.
be the two triangles the sides of which
all
*S^.
Denote the circular points on the by w, w.
tt
plane or plane of pro-
jection
Project
B
and
C
into twand w';
since the projection of
Further
A
»S^
.-.
S projects
into a parabola,
touches the line at infinity.
will project into the focus of the parabola, since
the tangents from the focus go through the circular points.
Using corresponding small
letters in the projection,
parabola goes through the focus, a,
a,
A, B,
.-.
h',
c',
C, A', B',
a
a,
co'
0),
C
on a
lie
circle.
on a conic.
of the above proposition follows at once
The converse
see
are concyclic.
b', c'
,
lie
we
whose sides touch a
that, since the circumcircle of a triangle
.•.
249
FOCI OF CONICS
CIRCULAR POINTS.
by
reciprocation.
We
253.
have in the preceding
article
obtained a proof of
the sides of two triangles touch
the general proposition that
if
a conic, their six vertices
on another conic by the projection
of
what
is
lie
a particular case of this proposition, viz. that the
circumcircle of a triangle whose sides touch a parabola passes
through the
focus.
This process
is
known
as generalisiiig by projection.
proceed to give further illustrations of
Let us denote the circular points
254.
n, n', and their projections on the course
co
and
to'
tt
We
will
it.
in the
plane by w,
are not the circular points in the
p
co'.
tt
plane by
Then of But
plane.
by a proper choice of the ir plane and the vertex of projection 0) and Q)' may be any two points we choose, real or imaginary. For if we wish to project H and H' into the points a) and &>' in space, we have only to take as our vertex of projection the point of intersection of the lines wD. and co'il', and as the plane TT some plane passing through o) and to'.
The
255.
figures in the
and
following are the principal properties connecting ji
and
tt
planes
when
Ci.
and
fl'
are projected into
&>
co'
1.
points 2.
Circles in the o)
and
co'
tt
Parabolas in the
the line ww' in the 3.
p
in the
tt
plane project into conies through the plane.
p
plane project into conies touching
plane.
Rectangular hyperbolas in the p plane, for which, as we H and ft' are conjugate points, project into conies
have seen, having
co
and
co'
for
conjugate points.
250
CIRCULAR POINTS.
The
4.
FOCI OF CONICS
centre of a conic in the
p
plane, since
of fin', projects into the pole of the line
Concentric circles in the
5.
having double contact at w and
A
6.
pair of lines
OA,
a>'
OB
it is
the pole
cow'.
plane project into conies
p
in the tt plane.
at right angles in the
p
plane
project into a pair of lines oa, oh harmonically conjugate with
This follows from the
Ota'.
Oft),
that
fact
OH, OH'
double lines of the involution to which OA,
0{AB,
therefore o{ah,
&)&)')
A
7.
120')
= -!
(§
82); from which
conic with
^
as focus in the
p
ft)
the two
and
256.
that
foci
&>'
S and
*S^'
plane.
of a conic in the j9 plane will project
In &)
by drawing tangents tt
plane.
importance that the student should realise
are not the circular points in the
tu'
they are the projections of
by
follows that
it
to the projection of the conic in the
It is of
and
ft)
the
plane will project into tt
into the vertices of the quadrilateral formed
from
are
belong, and
= — 1.
a conic touching the lines sw, sw' in the
And
OB
O
and
tt
plane
when
Xl'.
252 we have denoted the circular points in the tt plane w', but they are not there the projections of the circular
§
and
points in the
Our
p
plane.
practice has been to use small letters to represent the
So then we use to If H and H' are the circular points in the p plane,
and
wl for the projection of 12
and
fl'
respectively.
;
We
257. isation
will
now proceed
to
some examples
Consider the theorem that the radius of a point
of general-
by projection.
A
is
perpendicular to the tangent at A.
circle to
any
Project the circle into a conic through
C
of the circle projects into the pole of
The co'
CO,
at
generalised theorem
of a conic meet in
c,
is
251
FOCI OF CONICS
CIRCULAR POINTS.
to
and w
:
the centre
&)&>'.
that if the tangents at two points be any point on the conic and
and a
the tangent there
a{tc,
oi(o')
= —1.
258. Next consider the theorem that angles in the same segment of a circle are equal. Let AQB be an angle in the segment of which AB is the base. Project the circle into a conic through w and to' and we get the theorem that if q be any point on a fixed conic through the four points a, b, (o, w q (abuxo) ,
is
constant
Thus conies.
245).
the property of the equality
ment of a of
(§
of angles in the same segproperty
circle generalises into the constant cross-ratio
CIRCULAR POINTS.
262
FOCI OF CONICS
Again we have the property of the rectangular hyperif PQR be a triangle inscribed in it and having a right angle at P, the tangent at P is at right angles to QR. 259.
bola that
Project the rectangular hyperbola into a conic having a>'
for
eo
and
conjugate 'points and we get the following property.
If p
be
any iwint on a
jugate j)oints and ]) {qr,
coco')
k (pq,
coco')
260.
r
q,
= — 1 and = — 1.
Lastly
we
tivo
conic for which
w and
co
are con-
other jmints on the conic such that
if the tangent at
will generalise
p
meet qr in k then
by projection the theorem
that chords of a circle which touch a concentric circle subtend a constant angle at the centre.
CIRCULAR POINTS.
FOCI OF CONICS
253
Let PQ be a chord of the outer circle touching the inner and subtending a constant angle at G the centre.
The
concentric circles have double contact at the circular
points fl and H' and so project into two conies having double
contact at
The C, is
w and
centre
C
the pole of
&)'.
is
the pole of Hfl' and so
The property we obtain by If
c,
the projection of
woi'.
tiuo conies
projection
is
then
have double contact at two jJoints
if the tangents at these points meet in
c,
and if pq
of the outer conic touchinfj the inner conic, then
co
be
lo' and any chord
aiid
c (pqcoo)')
is
constant.
EXERCISES If
1.
be
centre of a conic,
tlie
12,
12'
the circular points at
and if Dim' be a self-conjugate triangle conic must be a rectangular hyperbola. infinity,
2.
If a variable conic pass
and touch two given straight
for the conic, the
through two given points F and F', shew that the chord which joins
lines,
the points of contact of these two straight lines will always meet
FF'
in a fixed point.
3.
If three
common
conies have
two points
in
common, the opposite
chords of the conies taken in pairs are concurrent.
and X, circumscribe the quadrangle ABCD. Two conies 4. in ^and G, Through A and B lines AEF, BGH&.VQ drawn cutting and S^ in F and //. Prove that CD, EG, Fll are concurrent. **>',
5.
If a conic pass
conic at a given point,
through two given points, and touch a given its chord of intersection with the given conic
passes through a fixed point. f2, 12' be the circular points at infinity, the two imaginary parabola coincide with 12 and 12', and the centre and second real focus of the parabola coincide with the point of contact of 1212'
6.
If
foci of a
with the parabola.
CIRCULAR POINTS.
254 7.
of
If a conic
two given
FOCI OF CONICS
be drawn through the four points of intersection and through tlie intersection of one pair of
conies,
common tangents, it also passes through pair of common tangents. 8.
a
Prove that,
common tangent
if
to
the intersection of the other
three conies pass through the same four points,
any two
of the conies is cut harmonically
by
the third. 9.
Reciprocate the theorem of Ex.
8.
from two points P, P' tangents be drawn to a conic, the four points of contact of the tangents with the conic, and the points P and P' all lie on a conic. 10.
If
[Project 11.
P
and P' into the
circular points.]
If out of four pairs of points every combination of three
pairs gives six points on a conic, either the four conies thus deter-
mined coincide or the four
lines
determined by the four pairs of
points are concurrent. 12.
Generalise by projection the theorem that the locus of the is the
centre of a rectangular hyperbola circumscribing a triangle nine-points circle of the triangle. 13.
Generalise by projection the theorem that the locus of the
centre of a rectangular hyperbola with respect to which a given triangle
is
self-conjugate
is
the circumcircle.
Given that two lines at right angles and the lines circular points form a harmonic pencil, find the reciprocals circular points with regard to any circle. 14.
to the of the
Deduce that the polar reciprocal of any circle with regard to any to the circular points as tangents, and has the lines from
point
the reciprocal of the centre of the circle for the corresponding chord of contact.
Prove and generalise by projection the following theorem of the circle circumscribing a triangle which is selfconjugate with regard to a parabola lies on the directrix. 15.
:
The centre
P
and P' are two points in the plane of a triangle ABC. BC such that BC and BA are harmonically conjugate with DP and DP' E and F are similarly taken in CA and AB respectively. Prove that AD, BE, OF are concurrent. 16.
D
is
taken in
;
17.
Generalise by projection the following theorem
:
The
lines
perpendicular to the sides of a triangle through the middle points of the sides are concurrent in the circumcentre of the triangle.
CIRCULAR POINTS. IS.
Generalise
:
The
FOCI OF CONICS
feet of the perpendiculars
from any point on the circumcircle are
of a triangle
255 on to the sides
collinear.
If two conies have double contact at A and B, and if PQ a 19. chord of one of them touch the other in Ji and meet AB in T, then
{PQ, 20.
RT)^-\.
Generalise by projection the theorem that confocal conies
cut at right angles. 21.
Prove and generalise that the envelope of the polar of a
given point for a system of eonfocals of the eonfocals
a parabola touching the axes
is
and having the given point on
its directrix.
system of conies pass through the four points A, B, C, D, the poles of the line AB with respect to them will lie on a line /. 22.
If a
Moreover
if
this line
conjugates of 23.
A
CD
and
I
meet
CD
in P,
the conic, and on the tangent at
X{PT, LL') = —
;
are hai-monic
L and
P
T
L'.
a point
prove that the locus of A'
meet a any point on taken such tiiat
to a conic
P
is
X
is
is
a straight
line.
Defining a focus of a conic as a i)oint at which each pair of
conjugate lines circle
1
PB
and
pair of tangents from a fixed point
third fixed tangent to the conic in
24.
PA
I.
is
orthogonal, prove that the polar reciprocal of a
with respect to another circle
the second circle for a focus.
is
a conic having the centre of
256
CHAPTER XIX INVERSION
We
261.
two
have already in
inverse points
'
'
§
13 explained what
with respect to a
meant
is
by-
being the
circle.
P and P' are inverse points if they lie on the same radius and OP. OP' = the square of the radius. P and P' are on the same side of the centre, unless the circle have an imaginary radius, = i/c, where k is real. centre of a circle,
As curve
P describes S and
S'.
a curve >S', the point P' will describe another 8' are called inverse curves. is called the
centre of inversion, and the radius of the circle radius of inversion.
P
If
is
called the
describe a curve in space, not necessaril}^ a plane curve,
then we must consider P' as the inverse of a sphere round 0.
That
is,
whether
P
P
with respect to
be confined to a plane or
OP
be a fixed point in space and P' be taken on
not, if
OP' = A constant
such
P' is called the inverse of P, and the curve or surface described by P is called the inverse of that described by P' and vice versa. that OP.
k",
,
It
is
convenient sometimes to speak of a point P' as inverse
to another point
that
is
P with
respect to
a
j)oint 0.
*By this
is
meant
the centre of the circle or sphere with respect to
which the points are inverse. 262. in
Prop. 'Phe inverse of a circle with a circle or straight line.
respect to
its jilane is
First let 0, the centre of inversion,
Let k be the radius of inversion.
lie
on the
circle.
a point
257
INVERSION
Draw Let
the diameter
P
OA,
let
be any point on the
OP. OP'
Then .-.
PAA'P'
is cyclic.
,*.
the angle
AA'P'
A' be the inverse of A.
P'
circle,
its inverse.
= Ic'=OA. 0A\
the supplement of
is
APP', which
is
a
right angle.
right angles to
,•,
J. 'P' is at
.•.
the locus of P'
A A'.
a straight line perpendicular to the
is
diameter OA, and passing through the inverse of A.
Next Let
not be on the circumference of the
let
P be
Let
OP
Let
A
any point on the
P'
circle,
circle.
its inverse.
cut the circle again in Q.
be the centre of the
circle.
Then OP. OP' = ^'•^ and OP.
OQ = sq,
of tangent from
•
Take
B
on
and BP'
is
parallel to
OA
•
to the circle
=
t"
(say).
qp;^^ ~ OQ
such that
OB OA
t''
k'
f
5
is
a fixed point
AQ. 17
INVERSION
258
BP'
, ^ And
.vY AQ .'.
OB = 7=nr = OA
h^ t,
>
^^
^ ^ constant.
t-
P' describes a
circle
Thus the inverse of the circle
is
round B.
another
circle.
Cor. 1. The inverse of a straight line through the centre of inversion. Cor.
is
a circle passing
If two circles be inverse each to the other, the
2.
centre of inversion
is
a centre of similitude
(§
25)
;
and the
radii
of the circles are to one another in the ratio of the distances of their centres from 0.
The student should observe that, if we call the two circles S', and if OPQ meet S' again in Q', Q' will be the inverse
8 and of Q.
Note.
The
part of the circle
*.S
which
corresponds to the part of the circle S' which
and
convex to
is
is
concave to 0,
vice versa.
Two
of the
common
tangents of
S and
S' go through 0,
and
the points of contact with the circles of each of these tangents will
be inverse points.
263.
point
is
Prop.
The inverse of a sphere with respect
This proposition follows at once from the figures
to
any
a sphere or a plane.
round
OA
as axis
will generate a sphere
;
last
by rotating the and line
in the first figure the circle
and plane each of which
is
the inverse
of the other; and in the second figure the two circles will
generate spheres each of which will be the inverse of the other.
INVERSION
Prop.
264. 0,
not in
its
For the
The inverse of a is a ciixle.
259
circle luith respect to
a point
plane,
circle
may
be regarded as the intersection of two
spheres, neither of which need pass through 0.
These spheres will invert into spheres, and their intersection, which is the inverse of the intersection of the other two spheres, that
is
of the original circle, will be a circle.
Prop.
265.
a point
in
its
A
circle will invert into itself with respect to
plane if the radius of inversion he the length of
the tangent to the circle
This
OPQ that
is
P
and
and Q are inverse
That
is,
part which
Cor.
1.
the centre
obvious at once, for
cut the circle in
P
from
if
of inversion.
OjT be the tangent from
Q, since
OP OQ = .
convex and
Any system
inverted into themselves
it
and
follows
points.
the part of the circle concave to is
OT-
inverts into the
vice versa.
of coaxal circles can be simultaneously if
the centre of inversion be any point
on the axis of the system. Cor.
2.
Any
three coplanar circles can be simultaneously
inverted into themselves.
For we have only
to take the radical centre of the three
circles as the centre of inversion,
and the tangent from
it
the radius.
17—2
as
INVERSION
260
Prop.
266.
Ttuo coplanar curves cut at the
any point
their inverses with respect to
Let
same angle as
in their plane.
P and Q be two near points on a curve
8,
P' and Q' their
inverses with respect to 0.
Then
since
OP. OP' .'.
.-.
at
QPP'Q'
.
OQ'.
is cyclic.
^OPQ = zOQ'P'.
Q move up to P so that PQ becomes the tangent to P; then Q' moves up at the same time to P' and P'Q^
Now S
= k' = OQ
let
becomes the tangent at P' .-.
the tangents at
P
The tangents however
Now.
if
to the inverse curve S'.
and P' make equal angles with OPP'. are antiparallel, not parallel.
we have two curves
»S\
and
PTi, PTo be their tangents there, and
*S^2
if
intersecting at P,
and
the inverse curves be
INVERSION /S/,
261
So intersecting at P', the inverse of P,
their tangents,
it
and PT/, P'T^' be
follows at once from the above reasoning that
^t^pt, = at,'P't:. Thus
and S^ intersect at the same angle
>S'i
If two curves touch at a point
Cor.
P
as their inverses.
their inverses touch
at the inverse of P.
Prop.
267.
P,
Q
If a
circle
8
a
he inverted into
inverses
P
of
and Q
^respectively,
circle S',
P' and
be inverse points luith respect to 8, then
and
Q', the
be inverse points luith
ivill
respect to 8'.
Let
be the centre of inversion.
Since
P
and
Q
are inverse points for 8, therefore
orthogonally every circle through circle
through 0, P,
Therefore
the
P and
Q,
and
8
cuts
in particular the
Q.
inverse
of
the
circle
OPQ
will
cut
8'
orthogonally.
But the
inverse of the circle
centre of inversion,
Therefore P'Q'
is
lies
OPQ
is
a line
the inverse of the circle
Therefore P'Q' cuts 8' orthogonally, that the centre of
Again,
since 0, the
OPQ. is,
passes through
8'.
since
orthogonally,
;
on the circumference.
it
8' orthogonally
every
circle
through
P
and Q cuts 8 P' and Q' cuts
follows that every circle through (§ 266).
INVERSION
262 Therefore,
if
^j be the centre of
S',
A^P'. J.iQ' = square of radius of
Hence P' and Q'
A
Prop.
268.
S'.
are inverse points for the circle S'.
system of non-intersecting coaxal circles can
be inverted into concentric circles.
The system being non-intersecting, the limiting points and L' are real.
L
Invert the system with respect to L.
Now L
and L' being inverse points with respect
to
each
the system, their inverses will be inverse points for
circle of
each circle in the inversion.
is
But L being the centre of the circle of inversion, its inverse Therefore L' must invert into the centre of each
at infinity.
of the circles.
Feuerbach's Theorem.
269. '
The
of inversion
principles
may
be illustrated by their
application to prove Feuerbach's famous theorem, viz. that the nine-points circle of a triangle touches the inscribed
and
the three
escribed circles.
Let
ABC
be a triangle, I
its
incentre and /j
ecentre
its
opposite to A.
Let
M and
ilfj
this ecircle with
Let the
Draw
line
be the points of contact of the incircle and
BC.
AII^ which
AL perpendicular
centre, orthocentre
Draw OD circle in
Now
bisects the angle to
BO.
Let 0, P,
A
cut
U be
BG
in
R.
the circum-
and nine-points centre respectively.
perpendicular to
BC
and
let it
meet the circum-
K. since
BI
and BI^ are
the
internal
and
external
bisectors of angle B, .'. .-.
.'.
since
inclined to
RLA BC
is
(§ 27,
(AR,II,) = -1, II,) = -1.
L(AR,
a right angle, Cor.
2).
LI and
X/j are equally
INVERSION the polars of
.-.
will
L
263
with regard to the incircle and the ecircle
be equally inclined to BC.
Now the
polar of
for the ecircle
Let
for
since
D
is
MiX
is
the incircle cutting
the polar of
MM,
(§ 12,
OD
in
X.
X
are
Cor.)
L
for the ecircle,
i.e.
L and
circles.
be the middle point of
tangent from
M and that
AXM,D = AXMD. aXM,D = zXMD.
conjugate points for both
N
for
the middle point of
.-.
Let
the incircle goes through
MX be the polar of L
Then
.".
L
through M^.
XL, then
the square of the
N to both circles = NX^ = ND^.
INVERSION
264
iV is on the radical axis of the
.-.
DM = DM,. ND the radical
two
circles
but so also
;
is
i) since
axis,
is
.'.
Now
the pedal line of
and
K
this is perpendicular to J/j.
goes through D, and clearly also,
K
is on the bisector of the angle A, the pedal since be perpendicular to AK.
.\
DN
the pedal line of K.
is
But the pedal
KNP
.'.
And .'.
is
since
JV
Now
line of
U
is
its
middle point.
circle.
invert the nine-points circle, the incircle and ecircle
whose centre
circle
N
is
and radius
ND
NL. The two
latter circles will invert into themselves
nine-points circle will invert into the line
BG
nine-points circle the inverse of that circle
D
UN=\OK.
the middle point of OP,
with respect to the or
K bisects KP. N
a straight line and
a point on the nine-points
is
must
line
and L, points on the
circle,
the inverse of the nine-points
But .*.
this line touches
the
nine-points
;
for
must be a
invert into themselves,
and the line, .*.
and
DL
is
circle.
both the incircle and
circle
;
N being on the
touches both
ecircle.
the
incircle
and
ecircle.
Similarly
CoK.
it
touches the other two ecircles.
The point
of contact of the nine-points circle with
the incircle will be the inverse of inverse of Mi.
M, and with the
ecircle the
EXERCISES 1.
Prove that a system of intersecting coaxal
inverted into concurrent straight 2.
A
sphere
is
inverted from a point on
to a system of meridians
two systems
and
parallels
its
surface
on the surface
of coaxal circles in the inverse figure.
[See Ex. 16 of Chap. II.]
circles
can be
lines. ;
shew that
will correspond
INVERSION
D
If A, B, C,
3.
265
be four collinear points, and
A', B',
C, D'
the
four points inverse to them, then
AC.BD A'C .B'D' AB.CD~ A'B' CD'' .
and
P
If
4. i*i,
be a point in the plane of a system of coaxal
P^, Pg
jfec.
of the system, Pi,
circles,
P"
P
If
5.
be
its
circles,
inverses with respect to the different circles
Pj &c. are concyclic.
P.-,,
be a fixed point in the plane of a system of coaxal
P
P' the inverse of
circle of the system,
with respect to a
the inverse of P' with respect to another
P" with
P'" of
circle,
respect to another and so on, then P', P", P'" &c. are concyclic.
POP', QOQ' are two chords
of a circle and Prove that the locus of the other intersection
6.
point.
POQ, P'OQ'
is
a second fixed
Shew that the
7.
circle of the
system
circle.
result of inverting at
a coaxal system
circles of
is
any odd number
of
equivalent to a single inversion at one
and determine the
;
a fixed
is
of the circles
which
circle
so equivalent
is
to three given ones in a given order.
Shew that
8.
BCD with (internally
the circles inverse to two given circles
P
be equal, the circle
and externally) the angles
given
circles.
9.
Three
of points
if
respect to a given point
circles cut
AB'C
touch at A.
10.
Prove that
if
of intersection of the
two
one another orthogonally at the three pairs prove that the circles through ABC,
CC
AA\ BB',
ACD,
PCD bisects
;
the nine-points circle and one of the angular
points of a triangle be given,
locus of the orthocentre
tiie
is
a
circle.
Prove that the nine-points circle of a triangle touches the inscribed and escribed circles of the three triangles formed by joining 1 1
the orthocentre to the vertices of the triangle. 12.
circles
The figures inverse to a given figure with regard to two and Cn are denoted by and S.. respectively shew that if
C'l
^S*!
;
C\ and Co cut orthogonally, the inverse of S^ with regard to C,
is
also the inverse of So with regard to Cj. 13.
shew
If A, B,
tliat
the triangles
Also that 0,
C
C
be three collinear points and
OBC, OCA, if
three other circles are
to cut the circles
any other
R of the three circles OAB are concyclic with 0.
the centres P, Q,
OBC, OCA,
angles, then these circles will
meet
circumcircle of the quadrilateral
drawn through
OAB
0,
A
;
0,
B
respectively, at right
in a point
OPQR.
point,
circumscribing
which
lies
on the
INVERSIOX
266
Shew that
14.
FCD;
and the
the circle
PAD
the circle
if
PAC
circle
PAR
cut orthogonally the circle
must cut the
PBG
circle
for obtaining the point
of contact of the nine- points circle of a triangle in circle
The
then
orthogonally.
Prove the following construction
15.
PBD;
cut orthogonally the circle
ABC
with the
:
HY is
contact
Y
drawn
BC in
A meets
bisector of the angle
tangent
The
to the incircle.
of this tangent
and
D
H.
From
H the otlier
line joining the point of
BC
the middle point of
cuts the
incircle again in the point required. 16.
Given the circumcircle and
the locus of the centroid 17.
A, B,
C
is
a
incircle of a triangle,
are three circles and
respect to any other circle.
Shew
that
a, if
c
h,
their inverses with
A and B
are inverses with
respect to C, then a and b are inverses with respect to 18.
A circle
radical axis of 19.
is
*S'
S and
Shew that
shew that
circle.
c.
inverted into a line, prove that this line
is
the
the circle of inversion.
the angle between a circle and
its
inverse
is
bisected by the circle of inversion.
The perpendiculars, AL, BM, CN to the sides of a triangle meet in the orthocentre K. Prove that each of the four circles which can be described to touch the three circles about AN, 20.
ABC
KM
KNBL,
KLGM touches the
circumcircle of the triangle
ABC.
[Invert the three circles into the sides of the triangle by means of centre K,
and the circumcircle into the nine-points
Invert two spheres, one of which other, into concentric spheres. 21.
22.
Examine the
where O the centre 23.
If A,
inverses of P,
particular case of
on
of inversion lies
Q be three Q with respect P,
lies
tlie
circle.]
wholly within the
proposition of
§
151,
*S'.
and if P', Q' be the P'Q' meet OA in ^,,. then
collinear points, to 0,
and
AP. AQ
if
OA'-
A.P'.A^Q' "
UI}
A circle is drawn to touch the sides AB, AG of a triangle and to touch the circumcircle internally at U. Shew that AE
24.
ABC
and the
line joining
opposite to
between
AB
A
A
to the point of contact
with
BC of the
ecircle
are equally inclined to the bisectors of the angles
and AC.
[Invert with
A
as centre so that
C
inverts into itself.]
267
CHAPTER XX SIMILARITY OF FIGURES
Homothetic Figures.
270.
If i^ be a plane figure, wliicli
of points typified
by P, and
we may regard
as an assemblage
be a fixed point in the plane,
if
and if on each radius vector OP, produced if necessary, a point P' be taken on the same side of as P such that OP OP' is :
constant (= is
k),
then P' will determine another figure F' which
said to be similar
Two
such
honiotlietic,
We
and similarly situated
figures
are
and the point
to F.
conveniently called, in one word, is
called their homothetic centre.
see that two homothetic figures are in perspective, the
centre of perspective being the homothetic centre.
Prop.
271.
The
line joining two jyoints in the figure
F is
parallel to the line joining the corresponding points in the figure
F' ivhich
is
homothetic with
it,
and
these lines are in
a constant
ratio.
P
and Q be two points in F, and P', Q' the correOP OP' = OQ OQ' it follows that and P'Q' are parallel, and that PQ P'Q' = OP OP' the
For
if
sponding points in F' since ,
PQ
:
:
constant
:
P'Q'
:
ratio.
In the case where
PQ
:
= the .-.
Q
is
in the line
constant ratio, for since
OP -.OQ - OP =0P'
:
OP it is still true that OP OQ = OP' OQ'
OQ'
:
-
:
OP'.
SIMILARITY OF FIGURES
268
.-.
.'.
F and F' be curves and *S" the tangents P and P' will be parallel. For the limiting position of the line through P
If the figures
Cor.
them
to
OP:PQ=OP':P'Q'. PQ.P'Q'=OP:OP'.
*S^
at corresponding points
the tangent at
P
is
and a near point Q on S, and the tangent at P' the limiting position of the line through the corresponding points P' and Q'.
is
Prop. The homothetic centre of two homothetic figures 272. determined by two pairs of corresponding points. For is
if
two pairs of corresponding points P, P';
the intersection of
Or
PP' and
Q, Q'
be given
QQ'.
where Q is in the line PP', is determined by the equation OP OP' = PQ P'Q'.
in the case
in this line
The point
:
is
:
thus uniquely determined, for
have to have the same sign, that
is,
OP
and OP'
have to be in the same
direction.
273. If
Figures directly similar.
now two
F
and F' be homothetic, centre 0, and its plane round through any angle, have a new figure Fi which is similar to F but not now figures
the figure F' be turned in
we
shall
similarly situated.
Two and
is
Two
such figures
F
and F^ are said
to be directly similar
called their centre of similitude.
directly similar figures possess the property that the
SIMILARITY OF FIGURES
Z POPi between points
P
and Pj
PQ:PiQi =
the is
lines
joining
constant.
Also
OP
269
two corresponding OP^ is constant, and
to :
the same constant, and the triangles
OPQ, OPjQi
are similar.
Prop. If P, P^; Q, Q^ be two pairs of corresponding 274. points of two figures directly similar, and if PQ, P^Q^ intersect in R,
is the
For since
other intersection of the circles P-RP^,
Z.OPQ=zOP,Q^
QRQj.
SIMILARITY OF FIGURES
270
OPR
.•.
Z
.-.
POP.R
Similarly
and Z OPiR are supplementary. is cyclic.
QiOQR
is cyclic.
Thus the proposition
is
is
proved.
Cor. The centre of similitude of two directly similar figures determined by two pairs of corresponding points.
been assumed thus
It has
Px nor with
If
far that
P
does not coincide with
Qj.
P coincide
with Pi, then this point
is itself
the centre of
similitude. If
P
coincide with Qi
we can draw
QT
and Qi?\ through Q
and Qi such that
z P,Q,T, then
T
= Z PQT
and Q,T,
:
QT= P,Q, PQ :
;
and 1\ are corresponding points in the two figures.
275.
When
two figures are directly
similar,
and the two
members
of each pair of corresponding points are on opposite
sides of
0,
and
collinear with
it,
the figures
antihomothetic, and the centre of similitude
is
may be
called
called the anti-
homothetic centre.
When
two figures are antihomothetic the line joining any P and Q, of the one is parallel to the line joining the corresponding points P' and Q' of the other hut PQ and P'Q' two points
;
,
are in opposite directions.
SIMILARITY OF FIGURES
Case of t^vo coplanar
276. If
we
271
circles.
divide the line joining the centres of two given circles
externally at 0, and internally at 0' in the ratio of the radii, it clear from § 25 that is the homothetic centre and 0' the
is
antihomothetic centre for the two
We
spoke of these points as
'
circles.
centres of similitude
'
before,
but we now see that they are only particular centres of similitude, and it is clear that there are other centres of similitude not
For taking the centre
lying in the line of these. circle to
A
of one
correspond with the centre A^ of the other, we
then take any point
P
may
of the one to correspond with any point
Pi of the other.
Let
The
S
be the centre of similitude
triangles
PSA, P^SA^
SA SA, = AP :
:
for this correspondence.
are similar, and
A,P,
= ratio
of the radii.
272
SIMILARITY OF FIGURES
Thus S
lies
Thus the
on the
circle
on 00' as diameter
(§ 27).
locus of centres of similitude for two coplanar
circles is the circle
on the line joining the homothetic and anti-
homothetic centres. This circle we have already called the circle of similitude and the student now understands the reason of the name. 277. If if
Figures inversely similar.
^ be
a figure in a plane,
a fixed point in the plane, and
another figure F' be obtained by taking points P' in the
plane to correspond with the points
OP
P
of
^ in
such a way that
and all the angles POP' have the same bisecting line OX, the two figures F and F' are said to be inversely similar is then called the centre and OX the axis :
OP'
is
constant,
;
of inverse similitude.
Draw P'L
perpendicular to the axis
OX and
in Pi.
Then
plainly, since
OX bisects Z POP' AOLP'=AOLP„ OP, = OP'.
and ,'.
OPi
:
OP
is
constant.
let it
meet
OP
273
SIMILARITY OF FIGURES
Thus the
figure formed
by the points Pj
will
be homothetic
with F.
Indeed the figure F' may be regarded as formed from a F^ homothetic with F by turning F^ round the axis OX through two right angles.
figure
The student that F', P.2
if
any
will
have no
OY be
difficulty in
taken through
if
typified
by P^
will
proving
himself
F and and produced to
F
to
formed with the points
be similar to F; but the two
similarly situated except in the case
for
in the plane of
P'K be drawn perpendicular that P'K = KP.2 then the figure
and so
line
OY
where
will
not be
coincides with
OX. 278.
If
P
and Q be two points
in the figure F,
and
P', Q'
the corresponding points in the figure F', inversely similar to
we
easily obtain that P'Q'
and we see that the angle regard to this last point directly similar
279.
we
see the distinction
and figures inversely
to
OP
it,
:
OP',
P'OQ').
In
of
between figures
similar.
in tivo inversely
find the centre and axis of similitude.
solve this problem
A. G.
PQ = the constant ratio POQ = angle Q'OP' (not
Given two pans of corresponding points
similar figures,
To
:
we observe
that
if
PP'
cxrt
the axis 18
274
SIMILARITY OF FIGURES
OX
in F, then
PF
FP' =
:
OP
:
OP'
since the axis bisects the
angle POP'.
PF:FP'=PQ:P'Q'.
.-.
Hence
if
P, P'
these lines at is
Q, Q'
;
F and G
in
be given, join PP' and QQ' and divide the ratio PQ P'Q', then the line FG :
the axis.
the
Take the point Pj symmetrical with P on the other side of axis, then is determined by the intersection of P'Pj with
the axis.
Note. The student who wishes for a fuller discussion on the subject of similar figures than seems necessary or desirable here, should consult Lachlan's Modern Pure Geometry, Chapter IX.
EXERCISES 1.
Prove that homotlietic figures
be projected into homothetic 2.
in
two
If P, P'
;,
Q, Q'
;
will, if
orthogonally projected,
figures.
R, R' be three corresponding pairs of points
figures either directly or inversely similar, the triangles
F'Q'R' are
sin)ilar in
PQR,
the Euclidean sense.
3. If S and *S" be two curves directly similar, prove that if S be turned in the plane about any point, the locus of the centre of
similitude of 4.
circle,
S and
.*>"
in the different positions of
S
will
be a
circle.
two triangles, directly similar, be inscribed in the same shew that the centre of the circle is their centre of similitude. If
Shew
also that the pairs of corresponding sides of the triangles
intersect in points forming a triangle directly similar to them. 5.
If
two
triangles be inscribed in the
same
circle so as to
be
inversely similar, shew that they are in perspective, and that the axis of perspective passes through the centre of the circle. G. If on the sides BC, CA,ABoi-e. triangle ABC points X, Y, Z be taken such that the triangle X YZ is of constant shape, construct the centre of similitude of the syltem of triangles so formed and ;
prove that the locus of the orthocentre of the triangle straight line.
XYZ
is
a
SIMILARITY OF FIGURES If three points
7.
ABC
opposite to A,
X^
T,
C
B,
Z be
taken on the sides of a triangle
and
respectively,
similarly situated ellipses be described round
common
they will have a 8.
The
275
if
three similar and
A YZ,
BZX and
CXY,
point.
circle of similitude of
two given
circles belongs to the
coaxal system whose limiting points are the centres of the two given circles. 9.
two coplanar
If
circles
be regarded as inversely similar, the
locus of the centre of similitude
and the axis
P
10.
and P' are corresponding points on two coplanar
regarded as inversely similar and
Q
this case.
when Q and similarity,
the 'circle of similitude,'
still
is
through a fixed point.
of similitude passes
is
S
is
circles
the centre of similitude in
the other extremity of the diameter through P, and
P' are corresponding points in the two circles for inverse
»S" is
the centre of similitude.
Prove that SS'
is
a diameter
of the circle of similitude. 11.
E,
AD
ABCD is a BC in F
and
cyclic quadrilateral ;
similitude for the circles on 12.
EF
prove that
AB,
CD
Generalise by projection
similitude of
two
circles is coaxal
is
;
AC
and
BD
intersect in
a diameter of the circle of
as diameters.
the theorem that the circle of
with them.
18—2
276
MISCELLANEOUS EXAMPLES Prove that when four points A, B, C,
1.
circle,
and that the
divided in the ratio
D
lie
on
BCD, GDA, DAB, ABC
lie
a circle, the
on an equal which joins the centres of these circles is of three to one by the centre of mean position
orthocentres of the triangles line
of the points A, B, C, D. 2.
ABC
is
the centre of
a triangle,
its
inscribed circle,
and
AB
Jj, 5i, G^ the centres of the circles escribed to the sides BC% CA, respectively Z, M, JV the points where these sides are cut by the ;
Shew that the orthocentres of the MC^A^, JVA^B^ form a triangle similar and
bisectors of the angles A, B, C.
three triangles LB^C^,
similarly situated to Aj^B^C^, 3.
ABC
is
and having
a triangle, Z,
,
Mj,
N-^
its
orthocentre at 0.
are the points of contact of
the incircle with the sides opposite to A, B, C respectively ; L.^ is taken as the harmonic conjugate of L^ with respect to B and C
and N^ are similarly taken P, Q, R are the middle points of MjM^, iVi#2- Again AA-^ is the bisector of the angle A cutting 3C in A^, and A^ is the harmonic conjugate of A^ with respect to B and C B^ and Cg are similarly taken. Prove that the line A.^BJJ^
M.2
;
ZjZg,
;
is
parallel to the line 4.
ABC
is
PQR.
a triangle the centres of whose inscribed and circum-
scribed circles are 0, 0'
;
Oi, 0^, 0^ are the centres of its escribed
and OjOo, Oa^s meet AB, BC respectively that 00' is perpendicular to LM. circles,
5.
If circles
in
L and
M
;
shew
be described on the sides Qf a given triangle as them having the inter-
diameters, and quadrilaterals be inscribed in
and one side of each through the middle point of the upper segment of the corresponding perpendicular, prove that the sides of the quadrisections of their diagonals at the orthocentre,
passing
laterals opposite to these
one.
form a triangle equiangular with the given
MISCELLANEOUS EXAMPLES
Two
6.
one so that
circles are
such that a quadrilateral can be inscribed in
touch the other.
its sides
277
Shew
that
the points of
if
contact of the sides be P, Q, B, S, then the diagonals of PQRS are at right angles and prove that PQ, ES and QR, SP have their ;
points of intersection on the same fixed line.
A straight line drawn through the vertex A of the triangle lines DF, DF which join the middle point of the base to the middle points E and F of the sides CA, AB in JT, Y shew that BY parallel to CX. 7.
ABC
meets the
\s,
Four intersecting straight lines are drawn in a plane. Reciprocate with regard to any point in this plane the theorem that the circumcircles of the triangle formed by the four lines are concurrent at a point which is coucyclic with their four centres. 8.
E and F are
9.
bola,
E
PE meets
and
two fixed points, P a moving point, on a hyperan asymptote in Q. Prove that the line through asymptote meets PF.
parallel to the other
through
Q
parallel to
Any
10.
and with
its
parabola
is
in a fixed point the line
described to touch two fixed straight lines
Prove that
directrix passing through a fixed point P.
the envelope of the polar of
P
with respect to the pai'abola
is
a
conic. 11.
8hew how
to construct a triangle of given shape
whose
sides sliall pass through three given points. 12.
Construct a hyperbola having two sides of a given triangle
as asymptotes and having the base of the triangle as a normal. 13.
A
tangent
is
drawn
to
an
ellipse so that the portion inter-
cepted by the equiconjugate diameters is
is
a minimum
;
shew that
it
bisected at the point of contact. 14.
A
parallelogram, a point and a straight line in the
same
plane being given, obtain a construction depending on the ruler only for a straight line
through the point parallel to the given
line.
Prove that the problem of constructing a triangle whose sides each pass through one of three fixed points and whose vertices 15.
lie
one on each of three fixed straight lines
is
poristic,
when the
three given points are collinear and the three given lines are concurrent. 16.
collinear
A, B, C, D are four points in a plane no three of which are and a projective transformation interchanges A and B, and
MISCELLANEOUS EXAMPLES
278
C
also
and D. Give a pencil and ruler construction for the point any arbitrary point P is changed and shew that any
into which
;
conic through A, B, C,
D
transformed into
is
itself.
Three hyperbolas are described with B,
17.
two common points P and Q ABC with P and Q for foci.
and that there
;
C
;
Shew
for foci passing respectively through A, B, C. is
C,
A; and
A,
B
that they have
a conic circumscribing
Three triangles have their bases on one given line and their Six lines are formed hy joining the point of intersection of two sides, one from each of a pair of the triangles, with a point of intersection of the other two sides of those triangles, choosing the pairs of triangles and the pairs of sides in Prove that the six lines form a complete every possible way. 18.
vertices on another given line.
quadrangle. 19.
problem
:
Shew that in general there are four distinct solutions of the To draw two conies which have a given point as focus and
intersect at right angles at
two other given
points.
Determine in
each case the tangents at the two given points.
A BO is inscribed in a circle of which two hyperbolas are drawn, the first has C as a focus, OA as directrix and passes through B the second has C as a focus, OB as directrix and passes through A. Shew that these hyperbolas meet the circle in eight points, which with C form the angular points
An
20.
is
equilateral triangle
the centre
:
;
of a regular polygon of nine sides.
An
21.
ellipse,
centre 0, touches the sides of a triangle
OA, OB, prove that AD, BE,
and the diameters conjugate D, E,
F respectively
22.
and
its
A
;
to
ABC,
meet any tangent CF meet in a point.
in
parabola touches a fixed straight line at a given point,
axis passes through a second given point.
envelope of the tangent at the vertex its
OC
is
Shew that the
a parabola and determine
focus and directrix.
23. Three parabolas have a given common tangent and touch one another at P, Q, R. Shew that the points P, Q, R ai'e collinear. Prove also that the parabola which touches the given line and the
tangents at P, Q,
R
lias its
axis parallel to
PQR.
Prove that the locus of the middle point of the common chord of a parabola and its circle of curvature is another parabola whose latus rectum is one-fifth that of the given parabola. 24.
MISCELLANEOUS EXAMPLES Three
25.
circles pass
A
intersections are A, B, C.
E
on the
are concyclic
F on
OCA,
circle
if
279
through a given point and their other point £> is taken on the circle OBC, the circle
OAB.
Prove that where
0,
.
.
AF, and so which must be taken. for the chord
D, E,
AF
AF.BD. CE = - FB DC EA,
F
stands
Also explain the convention of signs
on.
26. Shew that a common tangent to two confocal parabolas subtends an angle at the focus equal to the angle between the axes
of the parabolas. 27.
The
A,B oi& triangle ABC are fixed, and the foot A lies on a fixed straight line determine
vertices
of the bisector of the angle
the locus of 28.
A
straight line
AB
that the chord to
X at
A and
29.
On
AB, the
30.
ABC BE,
in
CF
cuts
two
X
and Y, so The tangents
fixed circles
oi
that P, Q, P,
S
lie
on a fixed
in four points
circle.
P Y are Shew that as P
a fixed straight line AB, two points
PQ
is
of constant length.
and XP, YQ meet line
ABCD
X is equal to the chord CD of Y. B meet the tangents to F at C and D
Shew
P, Q, R, S.
such that
;
6'.
in
a point P.
locus of i2
is
and Q are taken two fixed points moves along tlie which AB is an asymptote.
X and
a hyperbola of
A parabola touches the sides BC, CA, AB of a triangle Prove that the straight lines AD, D, E, F respectively. meet in a point which lies on the polar of the centre of
gravity of the triangle
ABC.
two conies be inscribed in the same quadrilateral, the two tangents at any of their points of intersection cut any diagonal 31.
if
of the quadrilateral harmonically. 32.
A
circle,
centre 0,
thi'ough
ABC.
The
P
.
points A", F' are constructed. 33.
inscribed in a triangle
is
on the circle meets BC in D. The line perpendicular to OD meets PD in D' The corresponding
tangent at any point
Two
Shew
that
AU
,
BE' CF' ,
points are taken on a circle in such a
are parallel.
manner that the
sum of the squares of their distances from a fixed point is constant. Shew that the envelope of the chord joining them is a parabola. 34. A variable line PQ intersects two fixed lines in points P and Q such that the orthogonal piojection of PQ on a third fixed
line is of constant length.
bola,
and
Shew
that the envelope of
find the direction of its axis.
PQ
is
a para-
MISCELLANEOUS EXAMPLES
280
With
35.
at
Show
to (A).
a focus of a given ellipse (A
P
any point
A
and
its
P
meets
TSD
\xi
BOY, COZ
OR
S
l)ethe
Shew
What
are the directions of
D its
circle passes
the sides BG, CA,
BY
and
through two fixed points.
in the plane of a triangle
AB
ABC, and X,
respectively, such that
Y,
Z
AOX,
CZ OQ and
If the points of intersection of
Q and
be respectively
R,
shew that
are equally inclined to OA.
The
middle points of the diagonals drawn, and the middle point of the intercept between any two sides is joined to the point in which they
39.
line of collinearity of the
of a quadrilateral
on
if
a rectangular hyperbola of which
is
are right angles.
AX
and AX,
that,
has a given focus and touches two fixed straight
ellipse
any point
is
points
ai'e
an
then the director
38.
Prove
fixed in direction.
is
?
If
37. lines,
*S'
are ends of a diameter.
asymptotes
and the tangent
described similar
it.
axis passes through a fixed point D.
further that the locus of
T
is
parabola touches two fixed lines whifh intersect in T,
focus, tlie bisector of the angle
and
as focus,
that (B) touches the minor axis of (A) at the point
where the normal at 36.
)
as directrix, a second ellipse (B)
it
Shew
intersect.
is
that the six lines so constructed together with
the line of collinearity and the three diagonals themselves touch a parabola.
The
40.
triangles A-^B^C^, A.>B./J^ are reciprocal with respect to
a given circle; B-^C^, C-^A^ intersect in P^, and B^C^, C^Ao in P.,. Shew that the radical axis of the circles which circumscribe the P^A.^B^ passes through the centre of the given
txnangles P^A^B.-,, circle.
A transversal
41.
in P, Q,
C
R
;
and
respectively in
cuts the three sides BC, CA,
P\ PQ'
.
QR' RP' .
=-P'Q
OC
a.
are bisected in A', B'
.
triangle
B
and
Q'R R'P. .
in the plane of a triangle
drawn a transversal cutting the OB,
oi
Prove that
Q', R'.
Through any point
42.
AB
also cuts thi'ee concurrent lines through A,
,
C
of the transversal are bisected in
ABC
is
The lines OA, and the segments QR, RP, PQ P', Q' R'. Shew that the three
sides in P, Q, R. \
,
lines A'P', B'Q', C'R' are concurrent.
43.
drawn
From any to a
point
P
on a given
given circle whose centre
is
circle tangents PQ, PQ' are on the circumference of the
MISCELLANEOUS EXAMPLES first
281
shew that the chord joining the points where these tangents first circle is fixed in direction and intersects QQ' on the line
:
cut the
of centres. 44.
any parabola be described touching the
If
triangle, the chords of contact will pass each
sides of a fixed through a fixed point.
From D, the middle point of AB, a tangent DP is drawn Shew that if CQ, CR are the semidiameters parallel to and DP, AB:CQ = 2DP:CR.
45.
to a conic.
AB
'
46.
The
side
BC of a
ABC is trisected at M, JV. Circles BC at M and AB at and AC at A'. If the circles touch
triangle
are described within the triangle, one to touch
H, the other to touch BC at iV one another at L, prove that CH,
ABC
BK pass
through L.
a triangle and the perpendiculars from ^, ^, C on the opposite sides meet them in L, M, xi respectively. Three conies 47.
are described
A
;
;
is
one touching
a second touching
third touching
at A, B,
C
AL,
tliey all
CX,
BM at
BM,
AL
CN at
at N,
M,
L and
N and passing
through
passing througli
J/and passing through touch the same conic. Z,
C.
B
;
a
Prove that
A
parabola touches two fixed lines meeting in T and the chord of contact passes through a fixed point A shew that the dii-ectrix passes through a fixed point 0, and that the ratio I'D to 48.
;
OA
the same for all positions of A. Also that if A move on a whose centre is T, then AO is always normal to an ellipse the sum of whose semi-axes is the radius of this circle. is
circle
49.
given the
Triangles which have a given centroid are inscribed in a
circle,
common
and conies are inscribed in the triangles so as to have centroid for centre, prove that they all have the same
fixed director circle. 50. is
A circle
is
inscribed in a right-angled triangle
and another
escribed to one of 'the sides containing the right angle; prove
that the lines joining the points of contact of each circle with the
hypothenuse and that side intersect one another at i-ight angles, and being produced pass each through the point of contact of the other circle with the remaining side. Also shew that the polars of any point on either of these lines with respect to the two circles meet on
drawn from any point form a harmonic pencil.
the other, and deduce that the four tangents
on either 51.
of these lines to the circles If
ordinates be
a triangle
PQR circumscribe a conic, centre C, and Q, R to the diameters CR, CQ respectively,
drawn from
282
MISCELLANEOUS EXAMPLES
the line joining the feet of the ordinates will pass through the points of contact of
PQ, PR.
Prove that tlie common chord of a conic and its circle of curvature at any point and their common tangent at this point divide their own common tangent harmonically. 52.
53. Shew that the point of intersection of the two comn)on tangents of a conic and an osculating circle lies on the confocal conic
which passes through the point
of osculation.
ABC, AL, BM, CiV are the perpendiculars on the sides and MN, jVL, LM when produced meet BG, CA, AB in P, Shew that P, Q, R lie on the radical axis of the nine-points Q, P. circle and the circumcircle of ABC, and that the centres of the circumcircles of ALP, BMQ, CNR lie on one straight line. 54.
In
a triangle
55.
A
circle
through the
of
foci
a rectangular hyperbola
reciprocated with respect to the hyperbola
is
shew that the reciprocal is an ellipse with a focus at the centre of the hyperbola and its minor axis is equal to the distance between the directrices of the ;
;
hyperbola.
If
56. A circle can be drawn to cut three given circles orthogonally. any point be taken on this circle its polars with regard to the
three circles are concurrent. 57. From any point tangents OP, OP', OQ, OQ' are drawn two confocal conies OP, OP' touch one conic, OQ, OQ' the other. Prove that the four lines PQ, P'Q', PQ', P'Q all touch a third con-
to
;
focal.
F
and
on two conies JJ Prove that the corners of the quadrangle whose pairs of opposite sides are the tangents at P, P' and Q, Q' lie on a conic which passes through the four points of intersection of f/'and V. 58.
and V
59.
P,
If
two parabolas have a
they cannot have a 60.
Q, Q' are four collinear points
respectively.
common
The tangents
those at A', B' in T'
;
real
common
self-conjugate triangle
focus.
to a conic at
two points A and
B
meet
in 7\
prove that
T(A'AB'B, = T'{A'AB'B).
A
moving in a plane always touches a fixed circle, moving circle from a fixed point is always constant length. Pi'ove that the moving circle always touches 61.
circle
and the tangent of
another fixed
to the
circle.
MISCELLANEOUS EXAMPLES
A
62.
system of triangles
is
283
formed by the radical axis and P to a coaxal system of
each pair of tangents from a fixed point
Shew
circles.
that
P
if
lies
on the polar of a limiting point with
respect to the coaxal system, then the circumcircles of the triangles
form another coaxal system. 63.
.Two given circles S, S' inteisect in A, B through A any is drawn cutting the circles again in P, P' respectively. ;
straight line
Shew
that the locus of the other point of intersection of the circles,
one of which passes through B, P and cuts S orthogonally, and the other of which passes through B, P' and cuts S' orthogonally, is the straight line through
B
Four points
lie
perpendicular to AB.
on a circle the pedal line of each of these with respect to the triangle formed by the other three is drawn 64.
;
;
shew that the four 65.
D
A, B, C,
AB
BC, CA,
in a,
drawn meet
lines so
are four points on a conic;
b, c
respectively
are harmonically conjugate to lines
Da,
that
a, /3,
66.
Db',
in a point.
a,
b,
AB
Dc meet BC, CA,
EF
and the conic in c
E
cuts the lines
and
F
a,
;
with respect to E, F.
in a,
/3,
y respectively.
b' c
The Shew
y are collinear.
Three
so that their respective diameters
circles intersect at
DO, EO, FO pass through their other points of intersection A, B, C and the circle passing through D, E, F intersects the circles again in G, H, I respectively. Prove that the circles AOG, BOH, C 01 are coaxal. 67.
A
conic passes through four fixed points on a circle, prove
that the polar of the centre of the circle with regard to the conic
is
parallel to a fixed straight line. 68.
The
triangles
PQP,
P'R
P'Q'L" are such that PQ, PR, FQ',
are tangents at Q, R, Q\ R' respectively to a conic.
Prove that
P{QR'Q'Ii) = P'{QJi'Q'Ji)
and P, Q, R, F, 69.
Q',
If A', B',
an involution, and CO', DD',
R'
lie
on a
conic.
C, D' be the /-•,
Q, R,
S
points conjugate to A, B, G,
D
in
be the middle points of A A', BB',
{PQRS) = {ABCD) {AB'CD'). .
70.
ABC
BDCX, CEAY, AFBZ be three (XBCD) (A YCE) {A BZF) = 1, and A D, BE, CF
is
ranges such 'that
a triangle.
If
ABC
.
.
be concurrent, then 71.
If
A'',
Y,
Z will
be collinear.
be a triangle and
line joining the circumcentres of
D
any point on BC, then
ABD,
ACD
(i)
the
touches a parabola:
MISCELLANEOUS EXAMPLES
284
the line joining the incentives
(ii)
bisectors of the angles
Find the envelope
72.
Two
of the line joining the centres of the circles
BD,
escribed to the sides
touches a conic touching the
ABC, ACB. (7/) respectively.
variable circles
S and
touch two fixed
a^S"
find
circles,
the locus of the points which have the same polars with regard to >S^
and
S'.
QP, QP' ai-e tangents to an ellipse, QMi?, the perpendicular on the chord of contact I'P and K is the pole of QM. If is the orthocentre of the triangle PQP', prove that UK is perpendicular to QC. 73.
H
74.
Two
touch one another at 0. Prove that the locus with respect to circles which touch the
circles
of the points inverse to
two given circles
is
another
circle
touching the given circles in 0, and
find its radius in terms of the radii of the given circles.
Prove that the tangents at A and C to a parabola and the meet the diameter through B, a third point on the paraHence draw bola in a, c, b, such that aB Bh = Ah bC ^ Bb cB. through a given point a chord of a parabola that shall be divided in 75.
chord
AC
:
:
How many
a given ratio at that point. there of this problem 76.
If A, B,
G
:
different solutions are
?
be three points on a hyperbola and the directions
of both asymptotes be given, then the tangent at
B may be constructed
by drawing through B a parallel to the line joining the intersection of BC and the parallel through A to one asymptote with the intersection of AB and the parallel through C to the other. 77.
A
circle cuts three given circles at right angles; calling
these circles A, B, C,
O
shew that the points where C cuts A and B touch O.
CI,
are
the points where circles coaxal with 78.
If
ABC,
DEF be
such that SD, SE,
SF
two coplanar
triangles,
cuts the sides BC, CA,
three collinear points, then SA, SB,
SC
and
AB
»S'
be a point
respectively in
cut the sides
FF, FD,
DE
in three collinear points. 79.
ABC
is
circle escribed to
way.
a triangle, i)
BC E \
and
is
F
a point of contact with are found on
CA,
AB
BC
of the
in the
same
Lines are drawn through the middle points of BC, CA,
parallel to
AD, BE,
the incentre.
CF respectively
;
shew that these
lines
AB
meet at
INDEX The
references are to pages.
Isogonal conjugates
37
Antiparallel
Asymptotes
36
103, 171 147, 153, 165, 205 Axes 101, 104, 107, 166 Axis of perspective 61, 62
Latus rectum 114, 126 Limiting points 21
Brianchon's theorem 215 Carnofs theorem 116 Ceva's theorem 33 Circle of curvature 120, 138, 158, 187 Circular points 242 Circumcircle 1, 5, 133, 245 Coaxal circles 20, 226, 234 Collinearity 31, Di, 214 Concurrence 33 Coufocal conies 165, 235, 236, 248 Conjugate points and lines 15, 74 Conjugate diameters 151, 155, 174,
Menelaus' theorem 31 Newton's theorem 117, 126, 136, 152, 177, 178, 179, 180 Nine points circle 3, 195, 262
•
Auxiliary circle
11(2
Conjugate hyperbola 170 Desargues' theorem 227 Diameters 106, 132 Director circle 150, 169 Double contact 246, 253 Ecircles 10, 262 Envelopes 130, 147, 212 Equiconjugates 156 Feuerbach's theorem 262
Focus and
directrix
10,
figures
113,
127,
145,
149,
164,
186 Ordiuates 107 Oithocentre 2, 133, 194, 232 Orthogonal circles 22, 73 Orthogonal involution 85, 94 Pair of tangents 111, 130, 147, 150, 168 Parallel chords 95 Parameter 136 214-
5, 133 Pole and polar 13, 92, 229 Projective propeities 45, 50, 53, 92
Quadrangle
91, 94, 96, 108,
249 54,
59, 60, 62
Homothetic
Normals
8
Pedal line
Harmonic properties 74, 75, 92 Homographic ranges and pencils
Incircle
24, 127, 137, 153, 179
Medians
Pascal's theorem
233, 246
Generalisation by projection
Loci
245 Signs 28, 119, 180 Similar figures 268 Simititude, centres of 24 Similitude, circle of 25, 271 Simson line 6
267
262
Inverse points 13, 256, 261 Involuti(m criterion 81 Involution properties 84, 93, 224, 227
222, 224
76,
Quadrilateral 75, 222, 224, 226, 248 Radical axis 17 Reciprocal figures 220, 237 Salmon's theorem 17 Self-conjugate triangles 16, 121, 236,
Subnormal 128 Symmedians 37 216,
Tangents
108,
127,
145, 164
Triangles in perspective
64
CAMBRIDGE J. L.
:
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