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SM - Conics Section - Besant

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SM - Conics Section - Besant

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UC-NRLF

5>
nJi't
SOLUTIONS OF EXAMPLES

CONIC SECTIONS,
TREATED GEOMETRICALLY

W. H. BESANT, D.Sc, F.R.S.


FELLOW OF ST JOHN'S COLLEGE, CAMBBIDGE.

THIRD EDITICN, REVISED.

CAMBRIDGE :

DEIGHTON, BELL AND CO.


LONDON : GEORGE BELL AND SONS.
1890
^,

PRINTED BY C. J. CLAY, M.A. & SONS,


AT THE UNIVERSITY PRESS.
PREFACE TO SOLUTIONS.

I HAVE frequently received requests for a book of


Solutions of the Examples in my treatise on Conic

Sections, but have never been able to find time to

prepare them.

Mr Archer Green, B.A., Scholar of Christ's

College, volunteered to undertake the task, with the


aid of my notes and his own, and, with the exception

of a few at the end, wrote out the solutions entirely.

Mr Green was however prevented by illness from

completing the revision of the proofs, and I am


much indebted to Mr J. Greaves, Fellow of Christ's

College, who kindly undertook to examine the rest of

the sheets.

The book will, I hope, prove useful both to

students and teachers, as a companion volume to

the treatise on Conic Sections.

W. H. BESANT.
Sept. 1881.

35784V
PREFACE TO THE THIRD EDITION.

The solutions have been revised,and many ad-


ditions have been made to them. They will now be
found to be in complete accordance with the sixth
edition of the Geometrical Conies.

W. H. BESANT.
Jan, 1890.
CONIC SECTIONS.

SOLUTIONS OF EXAMPLES.

CHAPTER I.

1. XF the tangent at P meet the directrix in Z, and 8


JL be the focus, PSZ is a right angle ;
.-. S lies on the circle of which PZ is diameter.

2. PN and QTIf be the ordinates at P and Q.


Let
Then PN QM SP aSQ XN XJf : :: : :: :
;

.'. the triangles PXN and QXM are similar and PX, ^X
equally inclined to XS.
3. Bv Art. 8, FS is the external bisector of the angle
PSQ.
4. SP : PK :: ^^ : ^A^ :: SE :
J5^A^;

EP bisects the angle SPK.


.-.

5. Since F, S, P and K lie on a circle,


the angle XSF= the angle FPX= the angle PTaS^

6. PiV^: P'N' :: /SP :


^K;
.\XK : XiV:: XX' : XN' ;

:, the angle XiV:A''=the angle XiV^X=the angle LN'N.


B. C. S. 1
2 Coriir, Sections,

7. ,Let Q be the point ^vhere the tangent at R meets


NP.
Tlien NQ NX : :: JSR : aS'X :: SA : ^X :: /S^P :
NX;
:,SP = QN
8. Let SY be perpendicular to the tangent at P and
G^Z perpendicular to SP,

Then, since the triangles PSY^ GPL are similar,


PG PL : :: SP :
aST,
or P(^ : SB :: /SP : SY.
9. If the tangent meet the directrix in Z, and be SP
drawn such that ZSP is a right angle meeting the tangent

inP,
then P will be the point of contact of the tangent ZP.
10. If P, Q be the extremities of the chord, and PK,
QL be perpendicular to the directrix,

SP PK : :: SA : AX :: SQ :
QL ;

.-. SP + SQ :: PK-\-QL :: .S'^ : ^X


Now the distance of the middle point of PQ from the
directrix is equal to half PK+ QL, and is therefore least
when SP + SQ is least, that is, when PQ goes through the
focus.

11. If TP, TP' be the fixed tangents, and the tan-


gent at Q meet them in E, E',
the angle PSE=i\\Q angle ESQ, and the angle QSE'=^i\iQ
axigleE'SP';
.-. the angle ^^^' = half the angle PSP\
}2. If perpendiculars from the given points PK, QL
be drawn to the directrix and S be the focus,

SP :
SQ :: PK :
QL, a constant ratio;

.•. the locus of /S' is a circle.


Conic Sections. 8

13. Let the normal at P meet the axis in G.

Taking as the fixed point in the axis, it is obvious that


the triangles OSR^ are similar ; GSP
.-. SO : SR SG :: : SP :: SA :
AX;
.'. SR is constant, and R lies on a circle of which aS' is

the centre.

14. AT AX : :: SR : SX :: SA :
AX;
.'.AT=AS.

15. aS'T' bisects the angle between SP and SQ, Art. 12,
and SR bisects the angle between QS, and SP produced,
Prop. II., Art. 5 ;
/. RST is a right angle.
16. The EA T, BRS are
triangles similar ;

/. AT SR EA ER : :: : :: AX :
SX;
.-. AT ^X : :: SR : aSX :: SA :
^A-^;

17. If TL be perpendicular to the directrix,


SR : 7X :: SA : AX :: SM : TL ;

.'.SM=SR.
18. FS\s the external bisector of the angle QSP, and
F'SofQSP';
.'. the angle i^^i^' = half the angle PSP\
19. Since the triangles SPN^ SGL are similar,
.-. GL : PN :: SG aS'P ^^ AX.
: :: :

20. If the normals PG, P'G' meet in Q, and QK be


drawn parallel to the axis to meet the chord in F,

Vq-, VPv.SG'.SPv.SA\AXv.SG''.SP'v. VQ VP' :


;

:. VP= VP\ or r bisects PP\


1—2
4 Conic Sections.

21. DS is the external bisector of the angle PSQ, and


ES ofpSQ;
.'. DSE is a right angle.
22. The semi-latus rectum is an harmonic mean be-
tween /S'Pand aS'P' ;

.-. 2SP SP'=SR PP\


. .

23. PE PL : :: PQ : P(y :: PV P^ : :: PP' :


2^P,
see Ex. 20 ;

:.PE : aS'/? :: SP' : .S'i?;

.-.P^^aSP'.

Similarly P'E=^SP,

24. The right-angled triangles DSQ, DSE have a


common hypotenuse.
Also SE^SR = SQ;
.'. the angle e^^=the angle ESP.

25. Let S be the focus and P and Q the given points.


Through P draw a straight line PK so that SP may bear
to PK the given ratio of the eccentricity.
Through Q draw a straight line QL so that SQ QL : in the
same ratio.
With centres P, Q and radii PA", QZ respectively describe
circles.

The perpendicular from aS' on a common tangent to these


circles will be axis.

26. Let the tangents at P and Q intersect in T.

Draw TN perpendicular to directrix and TM perpen-


dicular to SP,
Then SM TN : :: SA : AX.
But aST bears a constant ratio to SM, since angle TSM=
hnUPSQ;
:. ST bears a constant ratio to TN,
Conic Sections. 5

27. Let T be the intersection of the tangents at P


and p.
Draw TK perpendicular to Pp.
Then TK PL TP PG^ and TK
: :: : :
jt?/ ::
7> :
p^.

Again, draw GM, gm perpendicular to SP, Sp respectivelv,


and TN, Ta perpendicular to SP^ Sp respectively.

Then TP : PG :: TN MP : :: Tn :
mp ::
Tp :
pg;
.-. PL
TIsT : :: TIC :
pi ;

.'.PL=pL
CHAPTER II.

THE PARABOLA.

1. TiiE distance of the centre of the circle from the


fixed point is equal to its distance from the fixed straight
line, and therefore its locus is a parabola of which the fixed
point is focus and the fixed straight line directrix.

2. Through the vertex draw a straight line making the


given angle with the axis the tangent at the point where
;

the diameter bisecting this chord meets the curve will be


the tangent required.

Or, draw a radius vector from the focus, making twice


the given angle with the axis.

3. Since TA = AN, PN=2AY; .*. AY''=AS.AN.


4. Let jSY^ be drawn perpendicular to the line through
G parallel to the tangent.

Then in the right-angled triangles YST, Y'SG,


ST=^SG, and the angles YST, Y'SG are equal ;

.-. SY=:SY\
T), Y
Draw S perpendicular to the tangent and YA
perpendicular to the axis.
Produce aS'^ to JT, making AX equal to SA.
Then the straight line through X perpendicular to SX
is the directrix.

6. Let the circle touch the fixed circle in Q, and the


straight line in R\ let P be its centre, and S the centre of
tiie fixed circle.

Produce PR to M, making RM equal to SQ, then the


The Parabola. 7

straight line MX
drawn through M parallel to the given
line is a fixed straight line.

Then, since SP is equal to PM, the locus of P is a


parabola of which S is focus and directrix. MX
7. Draw SY^ SY^ perpendiculars on the two tan-
gents.

Then, SA be if perpendicular to YY\ A is the vertex.

Produce SA to X, making AX equal to AS X ;


is the
foot of the directrix.

8. If the tangent at the end of the latus rectum meet


PN in Q,
QN=XN=SP.
9. Since /S^FP and P^^^S^are right angles, P, N, S, Y
lie on a circle ;
.
TY,TP=TS.TN.
10. aS'^ is half TP,
and PT^=--PN^+TN^ = AAS.AN-\-AAN^',
SE'' = AN.XN=A]Sr.SP.
.'.

11. If aS'F be drawn perpendicular to the tangent and


A be vertex, SA F is a right angle ;

.*. A lies on the circle of which aS'F is diameter.

Draw -S'F perpendicular to the tangent, then if the


12.
circledescribed with centre S and radius equal to a qujirter
of the latus rectum meet the circle described on /S' as F
diameter in ^, is the vertex. ^
Produce SA to X, making AX equal to aS'^I, then X is
the foot of the directrix.

13. SN SP SN' SP\ : :: :

or AN-AS AN+AS AS-AN' AS+AN';


: :: :

.'. AN AS AS : :: :
AN';
.'. AN. AN = ASK
8 The Parabola,

Again, AAS . AN : A.AS'' :: A.AS'^ :


4:AS,AN'',
.'. PiV^ : SE" :: aS'T?^ :
P'iV'^
or FN : /S^ :. SR :
P'iV;
the latus rectum
.'. is a mean proportional between the
double ordinates.

14. Let V
be the middle point of the focal chord PSP',
and let the diameter through meet the curve in Q V ;

then, if QT, QM be the tangent and ordinate at Q, and


VL be ordinate of F,
VL = QM TM=SL) 2iii(}i

Vr- = QM^:=4AS .AM=2AS. TM=2AS SL.


.'. .

Hence the locus of F is a parabola of which S is vertex


and SL axis.

15. If P, P' be the given points, PK, P'K' perpendi-


culars on the directrix, the focus is the point of intersection
of a circle centre P, radius PK
with a circle of which P'
is centre and P'K' radius.

In general two circles intersect in two points, there-


fore two parabolas can be drawn satisfying the given
conditions.

16. If PG be normal at P, the triangles PNG, pPR


are similar ;
.-.
Pp : PN RP NG :: :
;

.-. i^P = 2iV6^ = latus rectum ;

.*.the locus of R
is an equal parabola having its vertex A'

on the opposite side of X, such that AA' i^ equal to the


latus rectum.

17. Let P, P' be the given points, aS' the given focus.

A common tangent to the circles described with centres


P, P and radii PS, P'S respectively will be the directrix.
18. If SP be the focal distance and /S'F perpendicular
to the tangent at P, Y lies on the circle of which SP is
diameter.
The Parabola. 9

Also the angle A Y"aS'= the angle SP F;


.*. A Y touches the circle.
19. The tangents at the ends of the focal chord PSP'
meet F on tlie directrix at right angles also the straight
in :

line through F at right angles to the directrix bisects PF'


in V;
:, FV^VP=VP'',
the directrix touches the circle described on
.'. PP' as
diameter.
20. Draw the farther tangent to the circle parallel to
the given diameter, then the locus of the point is a parabola
of which the centre is focus, and the tangent thus drawn
directrix.

21.Draw a straight line parallel to the given straight


line, on the farther side of it, and at a distance from it
equal to the radius of the circle, then the locus of the point
is a parabola of which the centre of the circle is focus, and

the straight line thus drawn directrix.

22. Let Q be the centre of the circle touching the


sector in R and AG in M.
Through C draw CB at right angles to AG, and on the
same side of it as Q, and draw perpendicular to theQN
tangent at B.
Tlien NQ+QM^BG= GQ + QR;
.-.
GQ = QN;
.*.
Q lies on a parabola of which G is focus and BN direc-
trix.

23. Y is the middle point of TP and Z of PG^ ;

therefore YZ is parallel to the axis.

24. If SQ be perpendicular to the normal PG,


PQ = QG,
and if QM be the ordinate, NM=MG ;

SM=AN and PN=2QM;


.'.

.-. QM' = AS. AN^AS. SM;


.*.
Q lies on a parabola of which S is vertex and SG axis.
10 The Parabola.

25. The triangle PSG is isosceles; therefore GL is

equal to FN.

26. If the circle described with centre S and radius


equal to the perpendicular from S on the tangent at P
meet the circle of which SP
is diameter in F, and the angle

SYA be made equal to the angle SP F, then the foot of the


perpendicular SA on YA will be the vertex.

27. Since SQ is double SA, ASQ (and likewise QSP)


is equal to the angle of an equilateral triangle ;

therefore SPand SQ are equally inclined to the latus


rectum.

28. QX^ = SX^ + SQ^ + 2SX. SQ


= 4AS'-+QG'-h2SQ.NG
=-4AS^+QN^ + 2QN.NG + NG' + 2SQ,NG
= 4AS^+ QN'' + NG^ + 2NG. SN
= 4^^=^ + NQ'' 4- 2 AN. NG
= 4AS^ + QN' -h PN' = 4AS^+ QP\

29. The angle


SPF= SPG - FPG = SGP - GPR= SHP ;

therefore the triangles SPF, SHP are similar ;

/. SF.SH^SP^=^SGl
SO. A, B, CyS lie on a circle ; therefore, if D be the end
of the diameter drawn through S, DA, DB, DC are per-
pendicular to SA, SB, SC respectively.

31. Since PQ
and PR
are equally inclined to the axis,
the circle through P, Q, R
touches the parabola at P;
therefore PQis a diameter of this circle.

Therefore PRQ^ the angle in a semicircle, is a right


angle.

32. Let MR and AQ meet in V.

Draw the ordinates VW, RZ.


The Parabola. 11

Then MW : MZ :: WV RZ AW . :: :
AN-,
:, MW AW : :: MZ AN; :

.'. AN AW:: AZ : : AN,


Again, VW' QN^ AW' AN^;: :: :

/.VW^ BZ^ AW AZ; : :: :

,\ V lies on the curve.

33. Let P, Q be the given points. Bisect PQ in F,


and draw FT parallel to the axis meeting the given tan-
gent P in T.
Draw PS, QS such that TP, TQ may be equally
inclined to the axis and to SP, SQ respectively. Pas', QS
meet in the focus.

Through P
draw a straight line PA" parallel to the axis,
making PK
equal to SP, then the straight line through K
at right angles to PK will be the directrix.
34. Let P be the vertex and Q VQ' be corresponding
ordinate.

Take M VP produced such that


in

QV'^4MP.PV.
Make angle TPS equal to the angle MPT, PT being
parallel to QQ, and make PS equal to PM.
S
Then is the focus, and the straight line drawn through
M at right angles to PM the is directrix.

35. Pil/2 :
QN' AM AN, QN being the ordinate
:: :

of Q;
.-.
A3f=4AN3ind3AM=4NM;
3AT=4QN=2PM.
.-.

36. Draw PiV perpendicular to AB.


Then AN NP CQ AC NP AC
: :: : :: :
;

PN^=AC,AN .\

Therefore the locus of P is a parabola of which A is


vertex and AB axis.
12 The Parabola.

37. The triangles LKP, PSK, KSA and TKA are


similar ;

.-. KL^ : /SP^ :: KP^ : KS'^ -^KA^ : AS^ ::TA AS :

::SP-AS AS. :

38. With centre S and radius one-fourth of the chord


describe a circle meeting the parabola in P. The chord
through S parallel to the tangent at will be the chord P
required.
39. PN-^ = 4:AS.AN=4AS.AN' + 4AS^
= P'N'^-^N'G'^ = P'G'-\
40. If Pp, P'p' be two parallel chords, and V^ V
their middle points, FF' is a diameter. Let FK^meet the
curve in Q,
Draw QT parallel to Pp^ then QT 'w> the tangent
at Q.

Produce FPto a point J/ such that Pr^ = 4Qil/. QV,


then the straight line drawn through at right angles to M
MV\9> the directrix.
Make the angle TQS equal to the angle TQM.
Then QS be made equal to Qi¥, S is the focus and
if

the straight line through aS' perpendicular to the directrix


is the axis.

41. Let the tangents at P and P^ intersect in T.

Then 4.SP .PV 4SP' P' V : .

::P'V' : PV"' :: TP^ : TP'^ :: SP : SP' ;

.-. PV^P'V.
42. If in the preceding Example P' 7" meets PV'm Z
and the sides of the triangle ABC are parallel to ZP, PT
and rZ respectively,
AB AG : :: TZ : TP :: Tr : TP.

43. If Uf rbe the vertices of the diameters bisecting


Pp, Qq.
PS.Sp :
QS.Sq :: ^^ SV : ::
Pp :
Q^.
The Parabola. 13

44. Draw R W, LZ parallel to QQ\


Then PU : PR^ :: LZ' : RW
:: QV^ : RW^ :: PV : PW :: PiNT :
PR-,
.-. PL'^=PR,PN.
45. This question is solved in Conies, Art. 212, p. 217.

46. PN^ : AN^ :: ^il/^ .

qj^s
.

.-. 4.AS : ^iV :: AM : 4AS.

47. Let AP, Ap meet the latus rectum in L and I

respectively.

Then PN"" : SL^ :: ^iV^^ .


^^2 ..
^^ .

^^^^ (Example 13)

:: PN^ :
pri";
.: SL=pn.
In like manner Sl = PN.

48. If PK^ QL be perpendicular to the directrix, and


QU to P^ produced, the angle SPQ=^i\iQ angle QPL' ;

PL'^SP = PK',
.-.

:, =
SQ QL^KL' = 2PK=2SP.
49. Is equivalent to Example 32.

50. Let Q be the point of intersection, and let QK be


the ordinate of Q.

Then AK QK PN NT
'. v. :
;

.-. 2AK,AN=QK.PN=PN'' = ^AS.AN;


.-. ^^=2^.S',
or C lies on a fixed straight line parallel to the directrix.

51. Let 75?, Tq be the fixed tangents, and let PQ touch


the curve in R.

Then SP^=:^Sp . SR = Sq SR = SQ'' .


;

.-. SP--^^Q.
14 The Parabola.

52. Let TMhQ the ordinate of T, and T^F perpendi-


cular to SP.

.'. the locus of 2" is a parabola of which X/S'is axis.

If Tir=2AS, TM'' = 4AS. XM, or X is the vertex.

53. Let the chord FQ meet the axis in 0, and the


tangent at A in V,

Then by Art. 48, VO^^ VP ,VQ\


:, F is a fixed point, and the locus of A is the circle
of which F is diameter.

54. Let the diameter Trmeet the curve in R.

Then the tangent at i?, being parallel to PQ, meets TP


at right angles in Z on the directrix.
Also TZ ZP TR RV
: :: :
\

.-. TZ^ZP,
Therefore yand P are equidistant from the directrix.
55. Let PTmeet the axis in U

Then PQ PT 1PV PT
: :: : :: "IPG :
Pt,
:: 2PN Nt : :: PN : AN.

56. If the tangents 7LP, TQ are equal, T lies on the


axis.

Let the tangent at R meet them in p and q.

Then, since T, jt?, ^ and S lie on a circle, the triangles


Sq r, SpP are similar ;

.-. Tq :
it?P :; TS : SP ;

.-. Tq=pP.
So TiP
= ^e.
The Parabola, 15

57. AN. NL = PN^r=4AS. AN;


.\ NL = 4AS,
But NG = 2AS;
.*. LG = half the latus rectum.
58. By Art. 5, P'S, Q'S are the external bisectors of
the angles PSA, QSA ;

therefore P'SQ' is a right angle.

59. The angles TCS, DRS are equal, being supple-


ments of equal angles SCP, SRC, Art. 35.

And the angle CTS=^ TQS=RDS;


.'. the triangles TOS, DRS ore similar;
.-. BR : TC :: RS SO RC GP: :: :
;

PC CT OR
:. : :: : RB.
Similarly TD DQ GR . :: : RD,

60. Let A D and XP intersect in Q, and let QM be the


ordinate.

Then QM DS AM : :: : ^.S' and QM PN XM XN; : :: :

.-. ^iJf : XM :: ^/S' : XN ;

:. AM : AS :: A>S :
^iV^;
.-. QM^ : PiV^2 ..
^j/2 .
^^2 ..
jij^ .

^jv,
or Q is on the parabola.

61. By Example 18, YY\ the tangent at the vertex, is


a common tangent.
SY^ = AS.SP, SY'^ = AS.Sp;
.'. YY'^=SY^ + SY'^=AS.Pp.

62. If P r be the diameter bisecting A Q,


AM=iAN.
16 The Parahola.

Also AM,MR = QM'^ = AAS,AM',


.-. MR = 4AS.
Now focal chord parallel to AQ
= 4:SP = 4:XN=-4AS + A3f=AR.

63. Let AR, CP meet in p.


Draw pN^pD perpendicular to CA^ CR, and let Dp meet
the tangent atA in M,
Cp CP
: CD Ci2:: : :: iVi? : CR :: ^i\^ :
AC-,
:. Cp = AN=^pM.
Therefore the locus of p is a parabola of which C is

focus and AM directrix.


64. If QMQ' be the common chord,
^AS'' = 4.AQ^ = AAM'' + 4.QM-=4AM'' + IQAM AS . ]

.-. ^iV/ishalf^AS'.

65. Let the fixed straight line ^^meet the tangent at


P in K.
Draw KY^ at right angles, and S Y^ parallel to KP.
Draw Y'A' perpendicular to the axis, and KL parallel to
BA.
Then, since A^F=^F; SA' = KL=^BA
'
therefore -4 is a fixed point.

Therefore KY^ touches the parabola of which S is focus


and A' vertex.

QQ. QD DR = QM^-DM'^ = Q3P-PN'^


.

= 4 AS, AM- 4 AS. AN=4:AS. PD.

67. Draw the double ordinate QMq; then, if the


diameter through Q' meet Qq in D',

QD\D'q = 4:AS.Q'D\
Tlie Parabola. 17

Now NT PN : :: QU :
QD' :-.
Uq :
4^^;
.-. D'q : \AS :: ^AN : PN :: PN : 2AS ;

.-. 2PN=D'q-=D'M+Mq = QM+QM\


Therefore the line through P bisecting QQ' is parallel to
the axis.
Hence the locus of the middle points of a series of
parallel chords is a straight line parallel to the axis.

68 \ Take CP, CQ two tangents such that PCQ is two-


thirds of a right angle ; join jSC cutting the curve in li\
and draw the tangent ARE. Then, Art. 38,
CSP^-CSQ = 120'', and CAR = \CSQ^m''',
.'. CAB is equilateral.

69. Draw AZ, AN perpendicular to the tangent and


SY respectively, and draw S3I perpendicular to ZA.
Then SM^ = AN^ = YN. NS= ZA AM, .

Therefore the locus of is a parabola of which A is vertex


/S'

and ZM axis.
70. If GZ be drawn parallel to PF and SZ to PG,
then SY, SZ are equal.

Therefore, ifZB be perpendicular to the axis, BS=AS.


Hence GZ touches an equal parabola of which B is

vertex and S focus.

71. If pqr be the triangle formed by the given straight


lines, describe a parabola passing through p, q and r having
its axis parallel to AS. (Ex. 45.)

If 5 be the focus of this parabola, draw SP parallel to


sp, PQ to pq, and to pr. PR
Then PQ :
pq :: SA : sa :: PR :
pr,
and the angles QPR, qpr are equal.
^
a parabola touch the sides of an equilateral triangle,
If
the focal distance of any vertex of the triangle passes through
the point of contact of the opposite side.
B. c. s. 2
18 The Parabola.

72. Let RWhQ. the ordinate of R.


Then
AN^ : ATV'^:: PN^ RW^ Pm QJ/2 : :: :
..
AN AM :
-,

.\AN AFT ATV AM, : :: :

or WN AN : :: MTV '.AW;
.'. RL :
QR :: AN AW PN '. v. -. NL.
73. Let PVhe the ordinate to the diameter RQM.
Then PiJf : i^iif :: PN 7W :

:: 2PN. AS : 4^aS' . ^iV^ :: 2^^ : PN ;

PM.PN=2AS.RM.
.-.

But PM'^ = 4AS.QV=4AS. RQ ;

.-. i^Jf :
i^Q 2PN Pil/ PP' :: : :: : P3f ;

.'.QM QR P'M Pi^. : :: :

74. Let PP
be the chord, TWV its diameter, RQM
the line parallel to the axis.

Then PM RM : PF TF PF 2Wr :: : :: :

:: 2VP.SW 4aS'^. ^F 2^^^ FV : :: :


;

PM.PV=2SW.RM.
.:

But PM.MP' = 4SW.QM,


.'. RM QJf 2PF i!/P, : :: :

or i?Q QM PJf MP\ : :: :

75. >S^i2, xS'r are the exterior bisectors of the angles


PSQ, pSQ respectively.
Therefore RSr is a right angle.
Therefore SD, which is half the latus rectum, is a mean
proportional between DR and Dr.
76. Let P
VP^ be parallel to the given straight line,
Q VQ' the chord joining the two other points of intersection
of the parabola and circle.
Let the diameters through F and meet the curve in V
p and p\
The Parabola. 19

Then pp' is a double ordinate ;


draw V'H parallel to
pp' to meet p V.
VV is perpendicular to QQ\ and therefore parallel to the
normal at p' ;

.-. VW :
p'g :: V'H :
p'n ;

.-. Vr = ^p'g.
77. The arcs Q U and R V are equal, since Q Fand UR
are parallel.

Therefore QR and ^F are equally inclined to Q V, that


is to the axis.

But QR and the tangent at P are equally inclined io the


axis ;

therefore UV\9> parallel to the tangent at P.

78. VR : VR' :: VR :
TQ' :: PV PP :

.'. VR. VR = QVK


79. If FR, QE meet the tangent at P in V and T,
TE EQ : :: Ti? : i^P :: PF : FQ. (Ex. 74.)
Therefore PPis parallel to TP.
80. IfQ be the vertex of the diameter bisecting the
chord Rr which meets the diameter PJVin W^

RW. Wr=r.4SQ.PiV.
Therefore the rectangle under the segments varies as
the distance of the point of intersection IV from P.

81. QSj Q'S are equally inclined to SP, and therefore


to the axis.

Therefore Q'S meets the curve at the end of the double


ordinate QMq, and, since =- AM. AM
AS^, the semi-latus
rectum is a mean proportional between and Q'M'. QM
Also, since the diameter through P bisects QQ', PS is an
arithmetic mean between QMsmd QM'.
2—2
20 The Parabola,

82. BB' will bisect CA' in V.


Let V be the middle point of B'B",
VV is parallel to the axis.
And BB" is parallel to VV, and therefore to the axis.

Similarly AA" and CG'^ are parallel to the axis.

83. Let (7 be the centre of the circle.

The angle between tangents to circle = PCP' = 2PSP'


=4 times angle between the tangents to the parabola.

84. The tangents at the ends of the focal chord PSP'


will meet in T on the directrix.
If the normals at P and P' meet in Q, TQ will be parallel
to the axis.

Let TQ meet the curve in ^? and PP' in V. Let QM be


the ordinate of Q.
Then XM= TQ = 2TV= 4Sp = 4Xn.
Therefore, if we take B in XM such that XB = 4.AS,
BM=4An, QM^=pn^ = 4AS,An = AS.BM.
Hence the locus of § is a parabola of which B is vertex and
BM axis.

85. Produce PA to P, making AP' equal to A P.


On AP^ as diameter describe a circle meeting the tangent
at P in T.

Join TA and produce to iV, making AN equal to A T.


In AN take a point S such that PN^ = 4:AS.AN, then S
is focus.

86. If G
be the intersection of the normals and Q vertex
of the diameter bisecting PSp,

ps.sp=AS. pp=As, tg:


87. If pq be a tangent parallel to P$, Tp=pP, and
T, ;?, qy S and lieon a circle.

Therefore the angles TSO, TpO are equal, and TpO is

a right angle.
The Parabola. 21

88. SM^ : AN' :: QM^ : PN^ :: AM : AN',


.-. SM'' = AM.AN
So ^^'2^^M'..4iV;
.-. MM' AN^MM' . .
(^il/-/SJ/') ;

.-. MM' SM-SM' : :: ^P :


^iV^;

89. If P, Q, P', Q' be the points of intersection, PQ,


P'Q! are equally inclined to the axis.
Hence the middle points of PQ and P'Q' are equidistant
from the axis.

Therefore, if P, Q be on one side of the axis and PQ'


on the other, the sum of the ordinates of and Q is equal P
to the sum of the ordinates of P' and Ql.

If P' be on the same side of the axis as P and (?, the


ordinate of Q! is equal to the sum of the ordinates of P, ^,
and P'.

90. Let the diameter through T meet the curve in W^


PQ in F, and PN
in t.

Let WZ be the ordinate of W \


draw Qq parallel to the
axis to meet PN.
QM.PN=PN.qN=tm-Pt''-^WZ'^-^AS, WV
= 4AS.AZ-4:AS.LZ=4AS,AL.

91. pX XA : :: PN : ^iV^ :: 4AS : PiV^

QM 4^.5'. ^Z. :: 4^.^. :


(Ex. 90.)
So qX XA PN ^X : :: :
;

pX-rqX XA PN+QM AL
.-. : :: :
;

.-.
^X+gX PN+QM X^ AL iX
: :: : :: : 7X.
But NP+QM = 2TL',
:. pX+qX=1tX,
or pt = ^^.
22 The Parabola.

92. Let TF, TD be drawn parallel to PE, QE normals


at P and Q.
The angle TFQ = PEQ = supplement oiPTQ=. TSQ ;

.'.
Q, S, F, Tlie on a circle.

Therefore TSF is a right angle.

So TSD is a right angle, and DF goes through aSI

93. \ipq be a tangent parallel to PQ, Tq^qQ.


Also, 7", j9, 2' and /S lie on a circle ;

therefore the angles Tpq, TSq are equal.


Therefore 2'Sq is a right angle.

94. Let i^O be the diameter through the given


point 0.
Take T in OR produced such that TR RO \ in the given
ratio.

If TP be a tangent, the chord POQ will be divided as


required. (Ex. 74.)

95. If QN be the ordinate,


BP + PQ = QN+BX-NX=BX+QN-SQ,
which is greatest when QN=SQ, that is when Q is on the
latus rectum.

96. If SZ and PG meet in Q siud QT be ordinate,


TA : ^^ QZ ZaS' QP PG 7W NG;
:: : :: : :: :

TN=2TA,
.-.

97. If Q F be the ordinate of the point of contact,


TP^PV,
Therefore the distance of V
from TQ is twice the distance
of P, or the locus of F
is a straight line parallel to TQ.

98. If TPSQ be the parallelogram, the angles TSP,


TSQ are equal;
therefore TPSQ is a rhombus and jTlies on the axis.
Therefore TSP is the angle of an equilateral triangle.
The Parabola. 23

99. If /S'Z, SZ' be the perpendiculars on the second


tangents, TQ, TQ' and PB be the common tangent, SY
perpendicular to it,

then angle A'S Y= AS Y= YSP ;

.-. A' lies in SP^ and A in SP' ;

.-. SP^SP".
Now SQ.sp= sr-=SQ' sr .
;

.-. SQ = SQ';
.-. sz=sz\
100. If the tangent meet ^F in F and the other
parabola in Q,

QM^ = ^AS,AM, AY^ = AS.AT,


QM AY=MT AT; : :

:. 2TM^=AT,A3L

This can be constructed by tsikmg AM


= 3fT, or by taking
AM=2AT, the two solutions corresponding to the two
points in which the parabola is cut by the tangent.
CHAPTER III.

THE ELLIPSE.

1. SD' = BC' = CS.SX,


Therefore CDX is a right angle.
2. ST, SP are equally inclined to PT, since pST is

parallel to S'P.
Therefore ST^SP.
3. PN : PG' :: SY SP ^C : :: : CZ>
:: PF AG : AG PG\ :: :

Therefore PN=AG.
4. T lies on a circle of which QQ^ is a diameter and F
centre ;

therefore VT= VQ.


Now QF2 : GP^-GV^ :: Ci)^ :
CP^,
or Kr^ CV.GT-GV^ (7i>^ CPl
.
:: ;

Therefore TV VG GD^ GP\ : :: :

Therefore GT GF GJ)'+GP^ GP\


: :: :

or (7^2 GF, GT AG' + BG^ C/P-^.


: :: :

Therefore GT^=AG^ + BG'\

5. Through T draw a straight line at right angles to


AA' meeting AP, A'P in JS, E'.

Then ET PN , v, AT AN \ :: GT-GA : GA-GN


The Ellipse. 25

Now CT CA : :: CA :
AN-,
/. CT+CA :CT-GA :: CA+AN CA-CN] :

^r PiV^ ^'T A'N ^T


.-. : :: : :: : PN,
Hence PT bisects any straight line parallel to ET termi-
nated by A'P, AP.

6. Draw CD parallel to the given line, and CP parallel


to the tangent at £>.

The tangent at P will be parallel to CD and the given


line.

7. SB : XE :: SA AX : ;: aS'T? : aS'X
Therefore XE=:SX,
and AT=AS.
8. Draw 6^Z perpendicular to aS'P.

Then PL = SB,
and aST : SP :: PZ PG^ : SR PG.
:: :

The angle aS'PaS" is greatest when SP Y is least, that is

when SY : iSP or BO CD : is least.

Hence SPS' is greatest when CD is greatest, that is when


CD = Aa
Hence SPS' is greatest when P is on the minor axis.

9. CE^ = CP' + PE^ + 2PF. PE = CD^ + (7P2


¥2CD PF=AC^-¥ BC^ + 2AC BC . .
)

:. CE=AC+BC
So GE=AC-BC.
{CP + GDf=AC^ + CB'' + 2CP CD, .

which is greater than {AC+BC)% since (7P CD is .


greater
than PP. Ci> or ^(7.^(7.

Similarly CP-CD is less than AC-BC,


10. Let /S"Q drawn parallel to SP meet the normal in
A^and/S^Fin^.
26 The Ellipse.

Then S'K^ S'P and KQ = SP ;

therefore S'Q = AA\


11. SYiS'Y' :: YP :PY' :: TP-TY: TY'-TP
::PG-SY:S'Y'-PG,
12. PS'Q is the supplement of QPS' + PQS\
and therefore equal to the excess of twice
is QPT+PQT
over two right angles,
that is, is the supplement of twice PTQ.
13. Since CZ and SP are parallel, the angle
CZP = SPY=SNY]
therefore Y", Z, (7, N lie on a circle.

14. Let ^ Q and /S'P meet in i^.


Then SA /Si?
: SG /S'P /S'^ :: : :: : AX.
Therefore R lies on a circle of which S is centre.

15. Since KPt is a right angle, t lies on a circle which


passes through S, P, S', K,
therefore GK SK : :: .S"^ : /S'P :: SA :
^X,
and aS^^ : tK :: /S'F : SP :: ^(7 : CD.

16. If /S'P meet S'Y' in Z, then since S'Y'= YZ,


SY' will bisect PG.

17. Let the circle whose centre is P touch the circles


whose centres are /S', in Q, i2. H
Then SP + PH= SQ-^QP + PH= SQ + HR.
Hence the locus of P is an ellipse of which S and H are
foci.

18. TN TO PN Ct. : :: :

Therefore TN.NG VT.NG PiV^ : :: : ^.PA^.


But PN^= TN.NG.
Therefore CT.NG^Ct. PN= CB\
The Ellipse. 27

19. TP TQ CD AC BC PF PG BC,
: :: : :: : :: :

20. PN^ AF.A'r 7W^ TA T^'


: :: : .

TN^ CT'-CA^ 7W CT :: : :: :

CT-CN CT CA^-CN^ C^^:: : :: :


.

.'. AF.A'F' = BCl


21. The perpendiculars from Ton /S'P, /S'Q, //P, HQ
are all equal.
Hence a circle can be described with centre T to touch
SPy SQ, HP, HQ.
22. If P, Q be two points of intersection,
PC bisects the angle ACa and QC bisects A'Ca.
Therefore PCQ is a right angle.
23. If aS'P, i7Q meet in i2,

PSQ + PHQ = 2PRQ - AS'()Zr- /S^Pi^,


and SQH+ SPH+ 2RQ T+ 2RP T= 4 right angles,
.-.PSQ + PHQ = twice the supplement of QTP.
24. Since t, P, aS', g lie on a circle,
the angle PSt = P^^ = STP.
25. Q'ilf : PM :: ^(7 : ^(7 :: PN :
QN.
Therefore Q'M : CN :: CM: QN.
Therefore QQ' passes through (7.

26. SY SP : :: BC : CD.
Therefore SY. CD = SP. BC.
27. If T be intersection of tangents at A and P,
then, since TC bisects AB, it is a diameter of the conic.
Therefore the tangent at C is parallel to AB.

28. The angles SPT, HPt are equal.


Also TP.Pt^ CD' = SP PH, .

or TP : SP :: HP : Pt.
Therefore SPT, HPt are similar.
28 The Ellipse.

29. PE=PE'=Aa
Therefore SE= HE\ and the angles SCE^ HCE' are equal.

Therefore the circles circumscribing SGE^ HCE' are


equal.

30. The angles KPG, are equal GPL ;


therefore KL
is a double ordinate of the circle of which PG is diameter.

31. If e be the centre, QN the ordinate, and T, T


the points where the tangent at P meets the tangents at
the vertices,

QN^ : SN.NH :: AT. AT : AH. AS :: BCT- : A'Sl


(Ex. 20.)

32. Since the tangents are equally inclined to SP, S'P


between them bisects
respectively, the bisector of the angle
SPS'j and therefore passes through the point where the
axis minor meets the circle.

33. If PQRS be the quadrilateral, p, q, r, s points of


contact, H the focus,

the angle pHP - PHs, pHQ = QHq,


SHr = SHs, rHR = RHq.
Therefore PHQ + SHE = PHS + QHR = two right angles.
34. SG : SO :: SP SY{see Ex.
:: 15)
:: CE EC PV
: :: ; VA.

35. The normals at P and Q will meet on the minor


axis in K.
Then angle between the tsmgexits
= PKQ^PSQ.

36. The a\ixiliary circle lies entirely without the ellipse


except at A and A' ;
therefore AA' is the greatest diameter.

The circle described on BB' as diameter lies wholly


within the ellipse ; therefore BE' is the least diameter.
The Ellipse,
*
29

37. Let any circle passing through N and T meet the


auxiliary circle in Q.

Then CN,CT=CA^ = CQ\


Hence CQ touches the circle at Q, and the circle cuts the
auxiUary circle orthogonally.

38. The angle PNY=PSY=PS'Y' = PNY',


Therefore PY PY' NY NY,
: :: :

39. PQ'' :
TQ2 :: SY.S'Y' : TY. TY.
But TQ'=TY.TY\
Therefore PQ" = SY.S'Y' = BC\
Therefore PQ^BC.
40. If QN and PM be the perpendiculars on the
given lines passing through R their point of intersection,
(7,

RN QN CP GQ\ : :: '.

therefore the locus of R is an ellipse of which the outer


circle is the auxiliary circle.

41. SP : S'P :: SY : S'Y' :: SY'' :


BC\
and S'Q :
SQ :: S'Z' : SZ :: BC^ : SZ\
Therefore SP.S'Q :
S'P.SQ :: SY'' : SZl
42. Let Cay Ch be the conjugate diameters, and Pm,
Pn ordinates of P.

Then Cm,CM=Ga\
and Cn^CN^Ch''',
.-. CM. Pm : Co" :: CW :
Pn.CN-,
.'. the triangle CPM varies inversely as the triangle
CPN.
43.

CA V : CPT :: CA^ : CT^ CN CT C/W CPT;


:: : :: :

therefore the triangles CA F, CPN are equal.


30 The Ellipse.

44. Let TPQ, Tpq be the tangents intersecting the


auxiliary circle in P, Q, p, q.
Let E, e be their middle points.

PE''-¥p6^ = ET^+ Te"- TP TQ^Tp . .


Tq

45. Let OQ' be a diameter equally inclined to the axis


with the conjugate to PP.
Then the circles described through P, P\ Q and P, P\Q'
will touch the ellipse at Q and Q\
Hence Q, Q' are the points at which PP' subtends the
greatest and least angles respectively.

46. Draw the tangent Qr.


Then, since the angles PSQ, QSr are equal, Q always lies
on the tangent at the end of the focal chord RSr,

47. The triangle YCY' will be the greatest possible


when YCY' is a right angle : P will then lie on the circle
of which SS' is diameter.
This intersects the ellipse in four points, provided SS'
is greater than BB\

48. The points where the lines joining the foci of the
two ellipses meet the common auxiliary circle are points
through which the common tangents pass.

49. The circle passing through the feet of the perpen-


diculars is the auxiliary circle of the ellipse.

60. Draw QiV perpendicular to ^^.


Then QN NA BP AP CA AT, : :: : :: :

and QN BN AT AB. : :: :

Therefore QN'^ AN. NB CA AB. : :: :

Therefore the locus of Q is an ellipse of which ^^ is major


The Ellipse. 31

51. PG, GN, NP are at right angles to CD, BR, RC


respectively.
Therefore the triangles CDR, PGN are similar.
Therefore PG CD : :: PN : CR :: BC : AC.
52. Let Pas', QS meet the elUpse and circle again
in^, ^.

And let P'Cjp' be the diameter parallel to SP,


Then, since pq isan ordinate,

SQ : SP ::
Sq :
Sp :: Qq :
pP :: AA' P'p\ :

Again, P.S'./S'/; : ^.S'.aS'^' :: CP'' :C^l


Therefore SR.Pp 2i^(7*^ Py2 j^/2^ : ::
.

or Pit? A A' Py^ ^^^2^


: :: ^

Therefore SQ SP w A A' Qq, \ :

and Qq = P'p',
5;^. If /S'P meet S' Y' in Z', SL' = AA';
therefore SR = AC.
54. Since the directions before and after impact are
equally inclined to the tangent at the point of impact, tlie
lines in which the ball moves will touch a confocal ellipse
or hyperbola.

55. Let the tangent at P meet the tangents at A and


^'inPandP^
Then, since the angles P.S'P, FSA and PaSP', F'SA'
are respectively equal, S (and similarly S') lies on the circle
of which FF' is diameter.

56. P', D\ the two angular points, will lie in PN, DM


respectively.
Therefore P'N : NC :: DM NC : :: DC : AC.
Therefore P' lies on a fixed straight through C. line

Similarly Q' lies on the other equi-conjugate diameter.

67. The angles aS'PaS", STS' are equal by Ex. 15.

A SPS' : STS' :: SP S'P , : ST.S'T :: CD"" : ST\


32 The Ellipse,

58. If T
be the centre, then, since the angles TSP,
TSA are equal, T
lies on the tangent at A,

59. Let QL be the ordinate of Q,


Then QL : LS :: CN NP, :

and QL : LS' :: CM : MD.


QL' : SL.SL' :: CM.CN PN,DM : :: ^(7^ :
BC\
or Q lies on an ellipse of which SS' is minor axis.

60. If P is the corresponding point on the ellipse, and


SZ the perpendicular on the tangent to the circle,
SZ AC CT-CS CT :: AC^-CS. CN AC^
: :: : :

:: aSP :
^(7;
.-. SZ=SP.
61. The tangent at Q is parallel to the normal at P ;

.*. the tangent at P


is parallel to the normal at Q.

62. If PQ P'Q' be the parallelogram, the angle DHE


is the supplement oi HPQ' + HQP'',
that is, of SPQ + SQP) that is, oi DSE.
Hence aS', //, i>, E lie on a circle.
63. Let the line through C parallel to the tangent
meet the directrices in Z, Z'.

Since the auxiliary circle is fixed, Y, S' Y' are fixed S


straight lines meeting
'
ZZ
in fixed points y y\

And Cy . CZ= CX.CS= CA\


Therefore Z and Z' are fixed points.
64. The angle S'TZ=STY=SZY=com^\QmQ\\i of
YZT,
Therefore FZand /S"Tare at right angles.

Qb. If G be the centre of the circle, GL bisects SP at


right angles.
Therefore SP is equal to the latus rectum.
The Ellipse. 33

^Q. If CZ be perpendicular to YY', the perimeter of


the quadrilateral is equal to SS' together with twice

CZ+ZY, which is greatest when CZ=ZYy that is when


SPS' is a right angle.

67. Draw SZ perpendicular to S'Z the straight line on


which aS" lies.

I^et PS'r be the chord parallel to SZ.


Produce PP' both ways to and M\ so that M
S'M=S'M' = AA\
Then the lines drawn through and M' perpendicular M to

SZ are fixed, and SP PM, SP' = P'M\


=
Hence the ellipse will touch two parabolas having S for
focus.

68. Let TQ, TQ! be tangents, Tthe middle point of

Then QV. FQ'=CP'-CV' = CF. FT.


Hence Q, Q', C, T lie on a circle.

69. Draw QM perpendicular to the minor axis.


Then AC^-CN^ BC^ AC'',
QG^ : :: :

or BG' AG'-GN'- AG\


QG-' : :: :

Therefore BG^-GN'-QN'' GN^ BG' AG^, : :: :

or BG'^-MG'' QM"" AG^+BG^ AG\ : :: :

Therefore Q lies on an ellipse of which BB' is minor


axis.

70. If QM be the ordinate of Q,


AM^ : GN^ :: QM^ : PN^ :: A3f,3fA' AG^-GN\
.

Therefore AM . AA' : AM^ :: AG"" :


GN\
or ^GN'^ = AG,AM.
But ^Q.^0 : GP^ :: ^J/.^C : C7iV« ;

therefore AQ,A0 = 2GP\


B. C. S. 3
34 The Ellipse,

71. SP : SN :: SO : CQ :: SC :
AC-QR.
Therefore SP .AG=SP .QE + SN, SO,
But /S'P : XS+SN :: /S^ : CA
or /S'P.^(7=XAS'.AS'(7+/S'iV^.>S'a

Therefore SP.QE^XS. SC=BG\


72. If the tangent meet the tangent at in T, and A
aS^'F, /S"Z be perpendicular to TS, and the tangent,
Z
T,Aj F, /S'', lie on the circle of which S'T is diameter.
The angles YTZ, ATS' are equal since ATS.S'TZ are
equal. Art. 68.

Therefore the chords FZ, AS' on which these angles stand


are equal.

73. If P, Q, P', Q' be the parallelogram, p, q, p\ q' the


points of contact, pq^ p'q' are parallel focal chords bisected
by PCP\
But QGQ' bisects pp\ qq' and is therefore conjugate to
PCP' and parallel to pq^ p'q'.
Therefore CQ = CQ' = CA
74. If T be the point from which the tangents are
drawn,
ST, S'T are perpendicular to TP\ TP respectively.
Therefore SP, S'P' are both parallel to CT,

75. CS'^ : CA'' :: CG GN GG XT


: :: : GN GT,

Therefore GG,GT=GS\
76. If PG be the normal at the point of contact,
GG,GT=GS\
Therefore 6r is a fixed point and P lies on the circle of
which GTi^ diameter.

77. Let the given straight line pq meet the axis t.

Let the tangents at p and q meet in Q.


Let GQ meet pq in V and the curve in P
Through P, Q draw P6^, Q(^' perpendicular to pq, meeting
the transverse axis in G and G\
The Ellipse. 35

Then CG' : CG CQ GP GP GV
:: : :: : ::
GT:Gt',
:.GG\Gt=^GG.GT=GS\
or G' is a fixed point.

78. Draw SY perpendicular to the tangent ; produce


SY io L making L F- YS.
The point of intersection of the circles described with
centres L and P', and radii A A' and AA'- SP respectively
will be the second focus.

79. Draw SY, SY' perpendicular to the given tan-

gents.
The point of intersection of circles described with
centres Y and Y\ and radius equal to GA will be the
centre.

80. If GS be drawn perpendicular to PQ, S will be


one focus.
If aS'P, PS' be equally inclined to GP and SQ, QS' to GQ,
S' will be the other.
Bisect SS' in G, and take GA in SS' such that
2GA = SP + PS'.
If Jl be the foot of the directrix,

GX,GS=GA-^.
81. Qq Aq PN AN, : :: :

and Er rA' PN NA'. : :: :

Therefore Qq.Rr :Aq. A'r PN"" AN. NA' :: :

v.BG'^'.AG'v.SL-.AG.
Now Aq :
qA' :: Aq^ :
Qcp,
and A'r : rA :: A'r^ : Rr^.
Therefore Aq.A'r :
Ar.A'q :: AG'^ : SLl
82. By Ex. 75. GT GS : :: GS :
GG;
therefore TS : GS :: SG : GG.
But TY PY : '.: TS : SG -,

3—2
36 * The Ellipse,

83. If be the intersection of the lines

84. TP : TQ :: Cr :
CQ\
and the angles PTQ, P'CQ! are equal.
Therefore PQ is parallel to P'Q',
85. /Si P, if,
/S' lie on a circle, and the triangles SCt^
P YS are similar.
Therefore St : Ct :: SP : SY CD BG, :: :

or St.PN : Ct.PN :: CD BG BC\ . :

Therefore St PiV^= (7i)


. . BG
86. If the tangent at Q meet the minor axis hi f,
the angle SQS^ = SfjS\ or ^^ is on the circle.
Now QM.Cf = BG' = PN. Ct.

Therefore QM PN : :: Ct : Clf :: Ct : Ct + St
:: BG BG+ CD by Ex.
: 85.

87. If S'Z be perpendicular to TY,


the angle >S'2'F= complement of TZY\ (Ex. 64)
- half supplement of YCY^ the angle at the
centre,
= GYY' and STY=S'TY\ ;

88. The tangents at L and L^ are perpendicular to the


tangent at P, and therefore and Z>' where they meet the D
tangent at P
are on the director circle.

Now DL DP : :: D'D DP; :

therefore PQ bisects the angle LPL\


Therefore LP + PL' - diameter of director circle.
89. ABfAE are equally inclined to BCy
and AB' = AD.AE,
Therefore AB is 2i
tangent.

90. If the tangent at P meet the tangent at A in T,


TS, TS' bisect the angles PSA, PSA.
The Ellipse. 37

91. If the chords of intersection NO, PQ 3neet in T


and CD, CE, C'D\ G'E' are parallel radii,

CD^ : (7^2 ..
TN, TO : TP TQ . :: CD"" : C'E\

92. IfPiVmeet (7i)in^,


PK PQ : :: SG : ^P :: .S'^ :
AX,
and PiV^. P^= PG PP= ^(72. .

Therefore PQ varies inversely as PiV.


93. Draw perpendiculars S Y, CE, S' Y' , SZ, CF, S'Z'
on tangents TP, TQ,
then CT'' = CF^+ TF^=^CZ^+TZ. = GA'^^SY. S' Y' TZ
= CA'^ + CB\
94. If the circle meet the minor axis in and L, the K
tangents at P
and Q meet either in jSTor L, see Ex. 15.

95. This problem is equivalent to Ex. 45.

96. Let CF bisecting the chord QSQ' meet the curve


in P and directrix in T
Let DCB' be the parallel diameter.
Then SR SG GS- GR GS GT- GV GT
: :: : :: :

:: GS^-GG"" SG^ SG. GS'


: :: : GS^
:: SP.PS' : GA'- :: CD' (7^^ ^^2 :
.. .
2)/) 2

by Art. 76.

97. Let the tangent at Q meet PN in P' and the axis


in U.
Then GT GN= GA' = GM. GU
.
;

therefore GT GM GU 6W, : :: :

or TM : 6'Jf :: iV^Z7 : GN.


But Q'M^ 2WPiV^. Q'Jf : :: :
TM,
or PN.Q'M GM.MU GN.NT : :: : GM.NU.
Hence PiV^ Q'ilf . : 6W. iV^r :: (7Jf ^iff . : (7Jf . PW
::
QM^:P'N,QM,
Now PiV"2 .
ON, NT :: BC'' : AG' ::^^2 ^^^^^2, .

Hence PN.Q'M : PiV^2 ..


q/j/2
.

p']S[,qM.
38 The Ellipse.

Therefore P'N : PN w AC \
BC,
or P' on the auxiliary circle.
is

98. The diameter bisecting PQ is fixed, hence V the


centre of the circle, is a fixed point.

VM bisecting RS at right angles, is a fixed straight


line ;

PQ BS are equally inclined to the axis


and ;

.'.CM and CV are equally inclined to the axis.


Therefore Jf is a fixed point, and RS a fixed straight
line.

99. The angle ^/S'a


= BAG+ SBA + SCA =BAG+ HBG+ HCB
= BAC+ supplement of BHG.
Hence if BHG is constant, BSG will be constant.

100. The angles SPT, HPt are each equal to SQH j

also STP^tQS-PtH
= HPt-PtH=PHt.
Therefore TP SP HP Pt, : :: :

or TP.Pt=SP.RP = GD\
Therefore GT^ Gt are conjugate.

101. GT bisects PQ and is parallel to SP ;

.'. T is the foot of the perpendicular from S' on PT.


See Cor. (3), Art. (66).

102. SG : GYis a given ratio and SY is fixed.


103. Take p a point near and let the focal chord p'Sq'
meet pq in O;
PQ :
p'q' .'.
pO.Oq :
p'O Oq . :: ST. Oq :
Sp' Oq'
.

:: ST.pq :
/SJ^ .
j^V ultimately.
The Ellipse. 39

104. Dropping perpendiculars from the focus on the


sides, their feet are the middle points, and, as they lie on a
circle, form a rectangle the diagonals, intersecting in H,
;

are therefore at right angles, and SAD can be proved equal


to HAB,
105. If CD, CP
meet the directrix in E
and G, ES is
perpendicular to the chord of contact of tangents from E,
which is parallel to CP.

106. If CP,DC meet the tangent at ^ in Q and R^


prove that AQ.AR = BC^=AS.AS\
Then QSA =ABS', and QBA = A QS\
CHAPTER IV.

THE HYPERBOLA.

1. If the circle whose centre is P touch the circles


whose centres are S and m Q and R, H
SP'-HP=SQ'-HR.
Therefore P lies on an hyperbola of which S and // are
foci.

2. SD'^=BC'= CS^- CA'^CS^-CX, CS= CS, SX.


Therefore the triangles SCD, SDX are similar.
,3. If the straight line meet the curve in P and the
directrix in F,
SF SX
: :: CS : CA SA:: : AX :: SB : SX.
Therefore SF=SE.
Draw PX perpendicular to the directrix.
Then PF PX SO CA SP
: :: : :: : PK,
Therefore SP^PF.
4. Draw SD^ SD' perpendicular to the asymptotes.
Then DD' is the directrix.

5. If the asymptote meet the directrix in Z>, then DS


drawn at right angles to CD meets the axis in the focus.
The Hyperbola, 41

6. If PK, QL be the perpendiculars from the given


points on the directrix PS—SQ-PK-QL wliich is con-
stant.
Therefore S lies on an hyperbola of which P and Q are foci.

7. If the circle inscribed in the triangle ABC touch


the sides in D, E, F-, D, (7, D
being given,
BA-CA=BF-EG=BD-DC.
Hence A lies on an hyperbola of which B and C are foci
and D a vertex.

8. FN :
Pg :: SY SP : :: BC CD :

:: Pi^ : ^(7 :: AC :
P^.
Therefore PN=AC,
9. Draw Ci) parallel to the given line and CP parallel
to the tangent at Z>.

Then the tangent at P is parallel to CD and the given line.


10. Let A'P and P'A meet in Q, and draw the ordinate
QM.
Then QM A'M FN NA\ : :: :

and QM ^3f PW NA. : :: :

Therefore QM' : AM, MA' :: PiV^ .


aN,NA'
:: ^(72 :
^C^,
or Q lies on an hyperbola having the same axes.

11. Let the tangent at P, AF and A'P meet the


minor axis in t, and E\ E
Then (7^ : FN :: C^ : ^iV^ :: CA.A'N :
AN,NA\
and Ci^' : PiV^ :: C^.^JV :: AN.NA'.
Hence CE-CE' : FN :: 2(7yP :: y4i\^. iV:^'.

Now P-.V2 : AN.NA' PiV. :: (7if :


^(72;

therefore C^- CE' = 2Ct,


Therefore Ft bisects every line perpendicular to AA'
terminated by A'F^ A P.
42 The Hyperbola.

12. SPTh an isosceles triangle since pST is parallel


to S'F.
Therefore SP = ST.
13. Draw JSD perpendicular to the asymptote and /SK
parallel to it.

If TS bisect the angle PSK, T being on the asymptote,


TP is the tangent at P. Draw aS'F perpendicular to it.

ThenCM which bisects i> F at right angles will meet


the asymptote in the centre G. DX drawn perpendicular
to CS will be directrix.

14. If the tangent at P


meet the tangents at A and A^
in Z and Z', ZS and Z'S are the internal, and external
bisectors of the angle ASP.
Hence the foci lie on a circle of which ZZ^ is diameter.

15. Draw P^ perpendicular to the directrix and DS


at right angles to the asymptote.
Draw cxs at right angles to the directrix meeting it in iv.
With centre P
and radius such that PS
PKwcs cZ), SP : :

describe a circle meeting Ds in S.


Then S is the focus.

16. Draw Qq', Pp and Rr, Qq parallel to the asymp-


totes.

Then GP :
GQ ::
Gp :
Gq' :: Gq : Gr :: GQ : GB.

17. QG^-GB^ : GN^-GA^ :: BG^ : AG^


:: PN^ :GN'^-GAK
Therefore PN^ = QB .
Q^'.

18. DN NA : :: QM MA : and ^iV^ : NA'


:: QJ/ : MA'.
Therefore ND.NE AN.NA' : :: ^3/2
.
^ jf . 3f^'
:: PiV2 : AN.NA',
or PN' = ND.NE.
19. ^, ^', F, Z lie on the auxiliary circle.

Therefore AT. TA'= YT . TZ.


The Hyperbola, 43

20. If the tangent at P meet the asymptotes in L


and L\ CD-PL = PL' \ therefore Q divides LL' in a
constant ratio.

Draw QH, QK parallel to the asymptotes.


Then QH. QK varies as CL CL' and therefore constant.
. is

Therefore Q lies on an hyperbola having the same asymp-


totes.

21. This is equivalent to the preceding.

22. Since SK= S'K, JTlies on the circle passing through


S, S' and P, and since KPt is a right angle, t lies on the
same circle.

Therefore GK : S'K SG SP
:: : :: SA :
AX,
and aS'^ : tK SY SP
:: : :: BC : CD.

23. Let P, P' be the points of trisection of the arc SS'


and let XM bisect aS'aS'' at
right angles, then SP = 2PM
^ndS'P' = 2P'M.
Hence P and P' lie on hyperbolas of which S smd S' are
foci and XM directrix.

If C\ a be the centres CS=4.CX and CS' = 4.CX.


Therefore C and C are the points of trisection of the chord
SS'.

24. Draw SZ parallel to the asymptote : the angle


STQ=TSZ=TSP.
Therefore SQ = QT,
25. Since the hyperbolas have the same asymptotes the
ratios CS : BG
CA are constant.
:

Let NP be the fixed line parallel to an asymptote, and


PQ proportional to an axis.
Then PQ'^ varies as CS^, that is, as CN .NP, that is, as
NP.
Hence Q lies on a parabola having NP for a diameter.
44? The Hyperbola.

26. PY.PY' = AC^-CP''=CD'^-BG'^=CS^


fSP--S'P\^
^y=(
27. Let TP meet the other asymptote in T'^ then
PT=PT\
Therefore PQ = R'P = QR.
28 Draw OrR parallel to PQ, meeting the ellipse and
hyperbola in r and R.
Let Oa, Oh be the axes, then since OP, Or are conjugate
in the ellipse, and OQ, OR in the hyperbola, if PN, QM,
rly RL be the or din at es,
0N.0b = rL0a', PN .0a = 0l.0h', QM.Oa
^OL,Ob', OM. Oh ^RL . Oa.
Therefore PN ON QM OM
: :: :

since rl 01 RL OL,
: v. :

or OP and OQ are equally inclined to the axes.


29. Through S draw SO parallel to the bisector of the
angle between the asymptotes meeting the asymptote which
is given in position in G.

Draw SD perpendicular to that asymptote, and DX to GS.


Then if A be taken in GS such that GA'^^GX.GS, A
is vertex.

30. If the tangents at P


and Q meet in T, then since
the perpendiculars from T
on SP, SQ, HP, are HQ
all equal, a circle can be described with centre to touch T
SP, SQ, HP and HQ,
3L If GL, GU be the asymptotes, S will lie in the
bisector of the angle LGL\
Draw PL, PH parallel to the asymptotes to meet them in
L and L' ;

and take S in GS such that GS^'^^GL GL\ .

then iS' i3 a focus.


The Hyperbola. 45

32. If the conjugate diameters PCP^, DOB' be given,


complete the parallelogram LML'M' formed by the
tangents at Z>, P, D' and P*.

The diagonals LL'^ MM are the asymptotes and the axes


bisect the angles LCM^ LCM'.
33. Let QT'and RQ meet the asymptotes in L and M,
Then QL : PR RH :: : 7Z :: CR : CT
:: RM T^ : :: PA" :
QM -,

therefore QL .
QM^ PH .
PA',
or Q is on the curve.

34. Let CL bisecting the angle ACE' meet PN in Z,


draw Q J/ parallel io LC.
Then CX is proportional to CN ;

therefore (7A . CM is proportional to CiV". NT+ CN"", that


is to CA\
Hence Q lies on an hyperbola of which CL and CB are
asymptotes.

35. Draw S Y perpendicular to the tangent and produce


it to Z making YZ-^SY.
Then if Q be the point of contact, and P the fixed point,
HP-PS=HQ'-QS=HZ;
therefore HP - HZ = PS,
or the locus of H is an hyperbola of which P and Z are
foci.

36. If P T, Pt the tangents to the ellipse and hyperbola


meet BC in T and t, then since the curves have the same
conjugate axis, for CA''= CS"" + GB^
Ct.PN=BC^=CT.PN,
or CT^CL
37. This problem is the converse of Ex. 3.
46 The Hyperbola,

38. If (r be the point of intersection

CG=i GP,
or G lies on an hyperbola having the same asymptotes.

39. The angle GYT = S'P Y' = S'NY' since S', P, N


and Y^ lie on a circle.

Therefore Y, Y\ C and N lie on a circle.

40. U SY meet .S^'P in Z, jSY=YZ; therefore S'Y


bisects PG.
Similarly jSY' bisects PG.
41. BC^ : ^6^2 ..
jsTG : ON :: CT.iVG^ : CN. CT;
therefore CT.NG = BCi
42. The angle STP = TS'P + S'PT= TS'P + SPT
= PS'S+ jSS't = supplement of PSt.

43. Pt.PT= CB" = SP S'P, .

or SP PT Pt S'P;
: :: :

and the angles SPT and tPS' being equal, the triangles
SPT, i^P/S'' are similar.

44. The circles SC^, B'GE stand upon equal chords


SG, S'G and contain equal angles SEG, jS'E'G, since CB is
parallel to the bisector of jSPjS\

45. If the tangent at P


meet the tangent at Aj the
vertex of the branch on which P
lies, in T, Tis the centre of
the circle inscribed in the triangle jSPS', since T/S, TS'
bisect the angles ASP, AS'P.
46. GT GA : :: GA : (7iV^ :: GP :
(7Q, or ^$ is

parallel to PT,
47. GE (7^ : :: ^/S^ : (7^ ;

therefore GE^ GS; but GD = GA.


Therefore AD and /S'^ are parallel.
48. If E be the centre of the circle and EK its radius,
EK GE : :: ^(7 :
SG,
The Hype^'hola. 47

or EK : CA :: BG SG+BG :

:: {SG-BG)BG : C^*
And SR' : BG ^C aST? BG;
CaS' :: : :: :

therefore ER' jSR GS-BG ^(7,


: :: :

or RR' : GS-BG aS72 ^(7 J5(7 GA, :: : :: :

Therefore EK=RR',

49. PM=^PL;
therefore GL = (ralf.

50. If Ga, Gb be the conjugate diameters and one


hyperbola touch the ellipse in P, the tangent at will P
meet Ga, Gb in T, t, such that TP = Pt = GD,
Hence PD is bisected by Gt,

and tD touches the other hyperbola and is parallel to GP.

51. If LL and MM' be the tangents,


GL : GM :: GM' :
GL\
or Zilf' and Z'iltf are parallel.

52. If the tangent at P meet the tangents at and A


A' in F and P' and QM be the ordinate of the centre of
the circle,
QM MS : :: .S'^ :
AF,
and QJf : MS' :: /S"^' : ^'P'.
Hence QM'^ : SM.MS' :: /S'^^
.
aF.A'F' :: aS'^^
.
^(72

(Art. 126).
Hence the locus of Q is an hyperbola of which S and S' are
Tertices.

53. If PM be perpendicular to the directrix,


PK : PJIf :: GS : (7^ SA :: : -4X :: /S'P :
PM,
or PK=SP.
48 The Hyperhola.

54. Let PD meet an asymptote in riy draw PI, Dm


parallel to Cm
Then Dm Dn = Pn. . PI ',

therefore Dn = Pn.
Therefore if LPL' is tangent at P, LD is tangent at Z>,
and (7P, CD are conjugate.
55. If QP meet the asymptotes in q and r, qQ = rR ;

therefore if PPe be the tangent at P,


CL : CN ::
qQ : Pe :: P^ :
gi2 :: CN : CJf.

56. If the circle intersect the axis in h, B,


CB,Cb=CS^
or CB^-\- CB,Bb = CA^-hOB^;
therefore CB.Bh = CA\
57. Let the straight line q'Q'APQq meet the asymp-
totes in Q\ Q.
Draw POP' parallel to ^P terminated by A'q, A'q,
Then PQ' = AQ^CP^ Qq,
and Q'q'=CP'=AQ' = PQ;
Pq = Pq.
'
therefore

58. T
is the centre of the circle inscribed in the

triangle PS'Q, therefore the difference between and PTQ


half PS'Q is a right angle.

59. Draw CD, CE parallel to OA, OB and PH, PK


parallel to and terminated by CE, CD.
Then PH OD : :: CK CE CP CA CB CP
: :: : :: :

:: CD CH OE P/r;
: :: :

therefore Pi7. PK= OD OE, .

or P lies on an hyperbola having CD, CE for asymptotes.


60. Draw PH, PK, QH\ QK' parallel to the asymp-
totes.

Then PL :
QM PH QH :: : :: QK' : PK :: QiV :
PR,
or PL.PR = QM.QN.
The Hyperbola. 49

61. If TK, TN be perpendicular to the directrix and


SP, TK=SN.
Therefore ST TK
: :: ST : /SiV a constant ratio,

and the angle between the asymptotes is double PST, that


is, double the external angle between the tangents.

62. Q'V^-RV^ = GD'^ = RV^''QV\


or QV^ + Q'r' = 2RV\
Again CT, CV= CP^ = CV. CT ;

hence CT= CT'


63. If r be the middle point of PQ, then since B, V
are the middle points of and LPQl^ LRU
is parallel RF
to the asymptote GUI.
Hence P3f+QN=2RB,
64. If TP, TQ be the tangents, PTQ, STS' have the
same bisector which passes through the point where the
circle meets BCB',

65. The tangents at P and Q intersect in t on the


circle and BCB'.
Hence the angle PtQ = PSQ.

66. The angle PNY= PSY= PS' Y' = supplement of


PNY\
67. The triangle YCY' is greatest when YGY' or
SPS' is a right angle.
In that case P 7" meets BC in t such that Ct = CS;
therefore CS. PN= Ct PN= BGK .

68. The triangle SPS' :


triangle StS' :: SP . PS' : St^

69. S'P is parallel to CFand S'Q to CZ.

Therefore S'T is parallel to bisector of YCZ and is

perpendicular to YZ.

70. If G be the centre of the circle, GL bisects SP ;

therefore SP = 2PL = 2SR,


B. c. s. .
50 TJ^e Hyperhola.

71. The tangent at Q is parallel to the normal at P,


therefore the tangent at P
is parallel to the normal at Q,

or CP is conjugate to normal at Q.

72. If F be the point from which the tangents are


drawn, SP and S'P' are both parallel to GY.

73. SC"" : AC"" CG CN CG CT CN CT,


:: : :: . : .

or CG.CT^SCl
74. By Ex. 73, G the foot of the normal is a fixed
point ;

therefore P lies on the circle of which TG is diameter.


75. If TP, TQ be the tangents, CT will bisect PQ in

and CT.CV=CT\
or PC is a tangent at F.

76. Let GQ meet the conjugate in G\


Then C(^' QG: :: CiV^ : NG :: ^C^ : 56^2,

Therefore, by Art. Ill, QG' is normal at Q.

77. If PM be drawn perpendicular to the directrix of


the parabola the angle PTQ = SPT-SQT=h8iU SPM-

78. If abed be the quadrilateral and S lie on the


circle the angle Hcd = Scb = Sab = Had,
or H is on the circle.

79.If PP' be the chord of contact and bisect CV


PP' then CF, PP' are parallel to a pair of conjugate
diameters in both conies.

Hence if from a common point Q, a double ordinate

QVQ' be drawn parallel to PP\ Q' must lie on both


curves.

Similarly RR' the line joining the other two common


points is parallel to PP\
The Hyperlola. 51

80. If SD, SD' are perpendiculars from the common


focus. on the asymptotes, DD' is the tangent at the vertex
of Pand a directrix of H,
If P be a common point, and PM perpendicular t'O

SP : PM :: SG :
CA,
but SP = PM+SX.
Therefore SP SX CS CS-CA CS AS.
: :: : :: :

Hence ^/S'. SP = SX. CS=BC^ = AS. SA \

or SP=A'S.
Therefore A'P touches the parabola at P,

81. With centre P, the given point and radius of the


given length describe a circle meeting the other asymptote
iwp.
Then pPQq is the line required.

82. Let CB, GA be semiaxes of the ellipse, Ga, Gb of


the hyperbola.

Let PN meet the asymptote in Q,


then QN"" : GN^ :: Gh^ :
Ga%
or QN^+GN^ : Ga^ + Gb^ :: CW^ .
^^2.

but SP + S'P = 2GA,


and SP-S'P = 2Ga.
Rence 4:GA.Ga=SP^-S'P^ = SN^-S'N^ = 4.GN.GS;
therefore (7^^
.
CS^ ... (7jv^2
.
(7^2
..
qn^ + cN'- :

(7a^ + C&2 :: GQ^ : (7>S^.

Therefore Qlies on the auxiliary circle of the ellipse.

83. Let Q be a common point.


Then SQ - QH=AA' and SQ-QP=SP- 2PH
=^AA'-PH,
Therefore QP = QH+ PH,
or Q must be the other extremity of the focal chord PH.
4—2
52 The Hyperbola,

84. li A'K meet the directrix in F^ then,


since SA' = 2A'X,
FA'S is an isosceles triangle and FS is parallel to KD,
Also A'F : FP :: AX XN : ^'/S'
:: SP :

or /^aS' bisects the angle ASP ;

tlierefore if SP and i>^ meet in Q, QSD is an isosceles


triangle.

Therefore Q lies on the circle of which A'D is diameter.

85. This problem is a particular case of Ex. 61.

86. PL.PL' = PL'=CD'^^PG.Pg',


therefore G, g, L, L' lie on a circle of which Gg is dia-
meter.
C is on this circle since GCg is a right angle.

The radius of this circle varies as Gg and therefore as CD


and therefore inversely as the perpendicular from C on
LL\
87. If PCP\ DGD' be conjugate diameters and Q any
point on the curve,
QP' + QP'' = 2CP^+2CQ^; QD'' + QD"'- = 2CD'^-¥2CQ\
Therefore
eP- + QP"" - Qlf^ - QU^ = 2CP'^- 2GD'^ = 2AC^'- 2BCI
88. If S^U, S'M' be drawn parallel to the asymptotes
LS\ MS' bisect the angles PS'L, PS'M\
Hence ZAS^'i^^half the angle between the asymptotes.

89. If Pr meets the tangent at A in F, T.S' bisects


the angle ASP\
therefore SP : ST :: PV VT AN: :: : AT.

90. If P
is a point of intersection, let the tangent and
normal of the ellipse at P
meet the transverse axis in T
and G, and the conjugate axis in t and g.
The Hyperbola, 63

Then, PT
being the normal of the hyperbola, the semi-
axes of which are A'C and B'C,
CT CN : :: SC^ :
A'C\ Art. Ill,

.-. AC^ CN^ SC A'C; : :: ;

.',CN,SG^AC.A'a
Again, Pg being the normal of the ellipse,

Cg : PN :: SC^ :
BG\ Art. 72,

and Cg,PN=EC\
.'. B'C^ PN' SC^ BC^
: :: :

and 'PN,SC=BC.BV.
Hence, if PN meet the asymptote in Q,
QN CN : :: B'C :
AV,
and it is easily deduced that

QN : PiV :: AC : ^01

91. Let ABCD be the quadrilateral. A, B, and (7

being fixed points.


Then AB+CD=BC+AD,
or CD-DA = CB-AB,
Hence D lies on an hyperbola of which A and C are foci.

92. Since Q, S, C, t lie on a circle, the angle

tQC=tSS' = SPt,
hence CQ is parallel io SY and CF, aS'Q are equally
inclined to SY-,
therefore SQ = CY=CA.
93. Draw SY perpendicular to the tangent and pro-
duce to Z making SY= YZ.
Then if P be the point of contact HZ= HP -SP = AA\
Hence the locus of ^ is a circle of which Z is centre.
94. RS and VS' bisect the angles PSQ and PS'Q;
let QS, S'P meet in Z.
54* The Hyperbola,

Then RSP + FS'Q = half PSQ + half PS'Q = half QSP


+ half SZS'-hsiU SQS'=QSP + hBlf SPS'-holf ^QS'
= QTP+ TQS-SQT=PTQ.
95. CgP is an isosceles triangle, and the angle
CGt = CPT^TCP;
therefore PG = Ct and CU' =-PG Pg=Cg .Ct= CS\ .

96.Since the asymptote bisects CD BA, CD is

parallel to the axis of the parabola and BA is parallel to


the other asymptote.
If QPVP'Q' parallel to BA meet CD in F,
er=rC' and PF=FP';
therefore QP = Q'P\
97. Let EL be the ordinate of E, and draw EF
perpendicular to PN.
Then, CD being conjugate to CP, the triangles CZ>J/ and
PFE are similar and equal.
/. CL = CN+ EF^ CN-\- DM,
.\CN'.CL','.AC:AC+BC,
and similarly EL PN :: BC BC- AC;: :

/. EL^ {BG-ACf :: PiV^^ BC^


: :

::CN^-AC^:AC^
CL' - {A C+ BCf
: : :
{A C+BCf.
PM be drawn from the centre perpendicular to
98. If
BC
AP PM PC PM, a constant ratio;
: :. \

therefore P lies on an hyperbola of which A is focus and


BC directrix.
If S be the other focus and SP meet the circle in Q
SQ = SP-PA = constant,
or, the envelope is a circle of which S is centre.
99. The conies will be confocal having their foci H and
H' on PG,
such that PH^=PT.Pt = CD\
For their locus see Ex. 9 on the ellipse.
The Hyperbola, 55

100. If SY, SZ, S' Y\ S'Z' be perpendiculars on tan-


gents at right angles
GT'^-GA''=-TY, TY' = SZ.SZ' = CL'K
If SYZ, S'Y'Z' are perpendicular to parallel tangents
and CWW be the perpendicular through the centre

2CW=SY^-S'Y'; ^CW' = SZ^S'Z\


and S Y- SZ=^ S' Y' + S'Z' ;

Hence CW^-CW'^=CB^^CB'\
101. The chord QR is inclined to the axis at the same
angle as the tangent at P and is therefore always parallel
to a fixed line.

102. TP and the asymptote subtend equal angles at S' ;

:. PS'T=S'TC=STP.
103. SF'- = FX"" + /S'X' = CF^ - (7X2 ^ ^X*
= CF^ + CS^-2GS.GX
= GF'^ + GS^-2GA'^ = GF^-GA'^ + GB^
= square of tangent from F
= FA.FB.

104. If S be the focus of the ellipse and S' of the


hyperbola,
GS : GA :: GA :
GS';
.'. aS' and S' coincide with the feet of the directrices.

The relation, GN. proves that S and S' are


GT=GA\
the feet of the ordinates, and the relation, Gt,PN-BC\
proves that t and t' are on the auxiliary circle.
Also Gif : GS=AG GS=GS' : :
Gt;
.*. the tangent intersects at right angles.

105. For PG.Pg=GD^, Art. 123, = PL'^- PL .PL\


and the diameter of the circle is Gg, which varies as CJJ,
Art. 123, and therefore inversely as GY,
CHAPTER V.

THE RECTANGULAR HYPERBOLA.

1. The angle PCL = (7ZP= complement of ZCF.


2. QV^=VP. Vp, hence VQ touches at Q the circle
QPp^

3. LP = PM=GD = PG = Pg.
Hence LGM is a right angle.

4. If X if be the straight line, and G be the corner of


the square, GL GM
is constant, hence
. touches an LM
hyperbola of which GL, are asymptotes. GM
5. Let AP^ A'P' meet in Q, and draw the ordinate
QM,
Then QM MA : :: PiV :
NA,
and Cif : MA' :: PW : iVL4'.

Hence QM^ : ^if/. JO' :: PN"^ : AN.NA' ;

therefore QM^ = AM. MA'.


Hence Q lies on a rectangular hyperbola having AA' for
transverse axis.

6. If P, P' be joined to Q meeting an asymptote in


R and B',
the angle

QRL = (7XP - QPL = LGP'-PP'Q= GR'P' = QEL,


The Rectangular Hyperbola. 57

7. Produce LP to M
making PM=PL, then if MG
be drawn perpendicular to the given asymptote CL^ C is
the centre. In CS the bisector of the angle take LCM
aS* such that CS is a mean proportional between CL and
CM.
Then aS' is focus, and X the middle point of CS is the foot
of the directrix.

8. The angle DCL^PCL,


and D'CL = P'CL'y
hence DCD' = PCP'.
9. A diameter is a mean proportional between the
parallel focal ciiord and AA', therefore focal chords parallel
to conjugate diameters are equal.

10. As in the preceding, focal chords at right angles


are equal, since diameters at right angles are equal.

11. If CD, Cd be conjugate to CP in the ellipse and


hyperbola,
(7i)2 ^ SP PS' = (7^2 = CP\
.

12. PN=CM) DM=CN',


and CD = CP.
13. SP.PS' = CD'' = CP^.
14. QV^=CV''-CP^=CV^-CT.CV=CV.VT.
Hence Q V touches the circle CTQ.
15. If i) be the intersection of tangents at A and B,
CD'^^AC'^BC^SCK
Hence D lies on the circle of which SS' is diameter.

16. If LPM^ G'PG be the tangent and normal to one,


LP = PM=CD = CP = PG = PG';
therefore GG' is the tangent to the second hyperbola.

17. The angle CRT=CQT+RTQ = 2CLQ + LTL'


= CLM+ CLT= CLE! + L'Cq = TEQ.
Hence C, T, R and R' lie on a circle.
58 The Rectangular Hyperbola,

18. QR^ = CN^=CA'' + RN^=CA^ + CQ^=AQ\


19. Let AQjAEhe the fixed straight lines, and P the
middle point of QOR.

Through G the middle point oi AG draw GH, (7A^ parallel


to AQ,AR,
and through P draw PHM, P^iV^ parallel to AB, AQ.
Then the complements AG, HK about the diagonal MN
are equal.
Therefore PH . constant, and P lies on a rectangu-
PK'i^
lar hyperbola, GH and GK for asymptotes.
having

20. Draw QB perpendicular to AB, and make AB


equal to GD, then ^ is a fixed point.
Then AD AB : BG GD QB
:: : :: :
DP,
or PD.DA^AB.BQ.
Hence P lies on a rectangular hyperbola of which AB is

one asymptote.

21. If Z>, ^, P, be the centres of the escribed and


inscribed circles,

OG,GF=DG.GE,
since the triangles OGE, DGF are similar.
Hence the hyperbola is rectangular since diameters at
right angles are equal.

22. If the diameters of the parallelogram LML'M'


meet in C, the angle SLS'^SL'S'.
Hence S and S' lie on a rectangular hyperbola circum-
scribing LML'M'. (Art. 137.)

23. If PSq, SQq be the chords and Z> be a point on the


directrix, such that DS bisects the angle QSp, then D will
lie in pq.
But DS is perpendicular to the asymptote since Pp, Qq
are equally inclined to it, therefore
D
lies on the asymp-
tote.

Similarly Pq, Qp meet in D', the foot of the perpendicular


from S on the other asymptote.
The Rectangular Hyperbola, 59

24. If V be the middle point of PQ and POP' be a


diameter,
the angle VNP = NP V= QP'P = PO V,
Hence O, P, F, N lie on a circle.

If OQ meet the given tangent in T, produce to V, OQ


making OV b. third proportional to and OQ with OT ;

centre F
and radius, a mean proportional between KO and
VT, describe a circle meeting the given tangent in its P
point of contact. In the tangent measure off

PL = PM=OP,
then OL and OM are the asymptotes.
25. Draw PD perpendicular to the base QR,
then, since PD'- DQ DB, --= .

DP is the tangent at P.

26. Let a circle on DE meet the hyperbola in P and Q,


draw the diameters PCP', QCQ\
Then, since the angle DPE either equal or supplemen- is

tary to DPE and DQE to DQ'E, the similar circle on the


other side of DE, will meet the curve in P' and Q'.

27. Let 0^2), 0^(7 be the fixed straight lines, PM,


PN, PL perpendiculars from the centre of the circle on
J5(7, AD and the bisector of the angle AOB.
Let PM, PN meet OL in m, n.
Draw X/, LI', Pr, Pr' perpendicular to BG, AD, LI, LV
respectively.

Then, BW^ 4- MP'^ = Am-¥ NP\


or Pm-PH'^^BM^'-AN''
which is constant,
P]Sr' = {Ll' + Lr'f = {Ll-Lrf + ALl'.Lr' = PM'' + 4.Ll.Lr,
Hence the rectangle LI Lr . is constant ;

but LI : OL in a constant ratio and Lr : PL is constant.


Therefore PL LO is constant, or P
. liies on a rectangular
hyperbola having OZ for an asymptote.
60 The Rectangular Hyperbola.

28. If P be the point of contact


CL = 2PN, CL' = 2CN;
therefore CL .CL'^2 Ca^ - 4.PN. CN.
Hence CN.CL : Oa^ :: Ca^ :
PN.GL,
or AC^ : Ca^ :: Ca^ : C£\
29. Draw CF conjugate to PQ,
Then, the angle

TPQ = PP'Q = PCV= GPQ' ;

therefore the angles CPQ, TPQ' are equal.

30. Let DB, DC meet the asymptotes in hand c :

draw AH, AK parallel to the asymptotes. Then OBb,


OAK are similar triangles, also OCc, OAK.
Hence
OB'.OA :: Oh : OK OH :: : Oc :: AK : Oc :: OA OC:
;

therefore A lies on the circle of which BC is diameter as


does D,
31. Let the tangents at P and Q meet the asymptote
in L and M.
The angle
PCQ = PCL - QGM^ PLC- CMT^ L TM
= supplement of PTQ.
32. Each hyperbola passes through the orthocentre of
the triangle ABC,
Hence D is that orthocentre.
Now the line joining the middle point of to the ^^
middle point of CD
is a diameter of the nine point

circle.

And ^ 52 +(72)2= square q^ diameter of circumscribed


circle.

Hence the circles intersecfc at right angles.

PN''=CN^-CA^=CN^-CN.CT=CN.NT.
33.

Hence the triangle GPN is similar to PTN, and therefore


iotTG
The Rectangular Hyperbola. 61

34. This problem is a particular case of Ex. 61 on the


hyperbola.

35. CM, CN which bisect PP\ PQ' are conjugate


being equally inclined to the asymptotes;
therefore PQ' is a diameter.

36. If AB he a diameter of the hyperbola, CD


subtends, at A
and B, angles which are both equal and
supplementary, and are therefore right angles.

37. If AQ\ BQ meet in R, the angle Q^Q' = supple-


ment of QBQ' = RBP',
therefore R hes on the circle.

38. Let CV, CV bisect PQ, PQ.


The angle PRQ^PQL= VCQ = CQ V,
so PR'Q=.CQV.
Draw VM perpendicular to CQ,
then PQ QR VM : :: :
CV,
and P'Q RQ VM
: :: : QF.
Now PQ = 2CV,
and PQ = 2QV',
therefore QR = QR\
39. Let PN meet CF in IT,

then CF varies as CG, and PF varies as PK, or Ct, or


CT.
Hence PF, FC is proportional to CG, CT or CSl
Hence P lies on a rectangular hyperbola having CF for
asymptote.
40. If the tangent at Q meet in V the line joining the
fixed points A and B,
VQ^=VA,VB.
41. If the chord QR meet the tangent at P in B,
RPL = QPE--^PRQ.
62 The Rectangular Hyperbola.

42. If D be the point,


CD.CT=2AC'^^ndLCD = Aa
43. CP = CD and are equally inclined to the asymptote.
44. Let BAD be the given difference, and draw CL
parallel to^Z>, meeting BA in X. CAL, CBL are similar
triangles ;
CL^=AL.BL,
45. If tQT, t'QT' be the tangents, and SM perpen-
dicular to the axis,

SQM=SQT-MQT=S'Qf-CtT
= QS'M+ QT'C- CtT= QS'M;
/. QM^=SM.S'M.
46. If V is the middle point of OP,
OTV= VOT, v DTP is a right angle,
= OCT, Art. 136;
'

/. O, F, (7, 7" are concylic, and


.-. VCT=VOT=OCV.
Similarly, if U is the middle point of OQ,
OTU= UTQ.
CHAPTER VI.

THE CYLINDER AND THE CONE.

1. Take two points E and A on the generating line,


and draw EX at right angles to the axis, making ^X equal
to^^.
Then the plane containing AX, and perpendicular to
the plane EAX will cut the cylinder in an ellipse of the

required eccentricity.

2. Take two points E


and A on the generating line
and with centre A, and radius twice EA, describe a circle
meeting EF in X.
Then the plane through AX perpendicular to the plane
EXA will intersect the cone in an ellipse of the required
eccentricity.

3. Take two points EA


on the generating line, the
least angle ot the cone will be, when EA
is double the

perpendicular from A on EF, that is when the semi-vertical


angle is equal to the angle of an equilateral triangle.

4. A tangent plane to a cone touches it along a


generating line OF, hence OF
is parallel to all sections

parallel to the tangent plane which are therefore para-


bolas.

If Cbe the centre of the sphere FES, the ratios


CS CA and CA
: CO are constant, and the angle OCS is
:

constant, therefore COS is constant, and S lies on a cone of


which is vertex and OC axis.
64 The Cylinder and the Cone,

5. Through the flames of the candles which are treated


as points, draw planes intersecting in the ceiling, in a
straight line, since these fixed planes must always be
tangent planes to the ball, the locus of the centre of the
ball ia a horizontal straight line.

6. The triangles A EX, A^E'X' have all their sides


parallel;

therefore
aS'^ : AX EA AX:: : :: E'A' : A'X' :: S'A' : A'X\

7. If C be the centre of the sphere FES, the angle


OCAS' is constant.
And the ratios CE :
CA, CE :
CV, and CS CA : are all
constant ;
therefore CS CV \^ constant and the
:
angle CVS is con-
stant, or VS is a fixed straight line.

8. Take two points E


and A on the generating line
and with centre A and radius AX, such that EA : AX in
the ratio of the eccentricity, describe a circle intersecting
FE in JT, the section of which is axis will have the ^X
required eccentricity.

9. XS, XS' are tangents to the same sphere FES of


which C is centre.

Let SS\ EF meet the axis in V\ L, and let CX meet SS'


in M.
Then CL,CV'^ CM. CX= CE\
Hence Fand F' coincide.

10. Draw CiV perpendicular to the axis,

then 2CN=A'D''-AD,
and 20N=0D+0D\
In CN take a point Q, such that
QN CN DO'' : :: :
DA^;
Tlie Cylinder and the Cone. 65

then since OD'-OD 2CN OD DA, : :: :

2QN OD'-OD DO DA
: :: :
;

also AD + A'D' 20N ^i> DO,


: :: :

and ^^'2 = Z>Z>'24-(^Z> + ^'i)7.


Hence QO^ : CA' :: 0Z)2 : DAI
Hence Q on a sphere of which
lies is centre, and there-

fore C lies on a spheroid, having OD for its axis, which is


oblate or prolate according as DO
is greater or less than

DA, that is according as the vertical angle of the cone is


greater or less than a right angle.

11. Draw a plane CT through C, the centre of the


sphere perpendicular to the axis intersecting the tangent
plane at S in T. Then since the angles COB and CTS are
equal, and CB= CS, it follows that CT= CO.
Hence S lies on the surface generated by the revolution of
the circle of which CT
is diameter.

12. Only two circles can be described passing through


Sj and touching the generating lines in which the plane
through S and the axis intersects the cone.
If ST, ST' be the tangents, and OS meet the circles in D
and D'j
the angle DST^^ half SCD = half SC'D' = D'ST.
Hence the planes of the corresponding sections make equal
angles with OS.

13. If the plane of section meet the plane through O


perpendicular to the axis in the hne ZZ', and be per- PK
pendicular to ZZ'
OP ov OQ bears to P^a constant ratio.
Hence if P' in the projection corresponds to P
OP P'K in a constant ratio.
:

But OP' is equal to the perpendicular from P on the axis


which bears to OP a constant ratio.
Hence the ratio OP P'K: is constant.
B. C. S. n
66 The Cylinder and the Cone,

14. A'Q — QA = SS' and a right cone can be constructed


of which Q is vertex, such that the generating lines inter-
sect the ellipse.

Hence
PQ + AS=AS'¥QR + RP = AE+EQ + SP=AQ + SP.
15. Through the vertical straight line which is the
locus of the luminous point draw vertical tangent planes to
the ball intersecting the inclined plane in O Y, OZ. The
locus of G
the centre of the shadow will be the straight
YZ, at right angles, SY, SZ being perpendi-
line bisecting
culars from the point of contact of the ball on OY, OZ
which are tangents to the elhptic shadow.

16. The given plane to which the sections are perpen-


dicular must be supposed to contain the axis of the cone.
TakeOK on the generating line equal to AS.
Draw KG perpendicular to OK^ meeting a line through
O at right angles to the axis in G, then (r is a fixed
point.

Draw GM perpendicular to AS,


then CE : EO :: OK KG,
:

or KG KA : :: AE : CE.
Hence the angle KAG^ACE^h^MKAM.
Hence AS touches a circle centre G and radius GK
17. VP=VQ=VA+AQ = 2AE+AN=2AS+AN.
18. EX\ X'A' will be parallel to EX, XA.
Hence AF be drawn parallel to A'E\ the ratio of
if the
eccentricities AE AF which constant.
is : is

19. The volume varies as the area A VA' and BC;


and the area AVA' varies as ^F. VA', or AD .
A'D',
that is, BG\
Hence BG is constant.
The Cylinder and the Cone, 67

20. A'O -0A= A'E' -EA= A'S- SA = SS'.


Hence the locus of is an hyperbola of which A and A'
are foci.

21. B<J'^=EC.CE\
Hence the locus of C is an hyperboloid of revolution
generated by the revolution round the axis of the cone of
an hyperbola of which OE, OE' are asymptotes.

22. CA and EA are the bisectors of the angle OAS.


Hence the sphere on CE as diameter, intersects the plane
of section in a circle of which AA' is diameter.

5—2
CHAPTER VII.

CURVATURE.

1. SL being a fourth of the chord of curvature at L


through S, LG the normal is a fourth of the diameter of
curvature.

2. The normal and tangent at L are equally inclined


to the axis.

3. Draw SO at right angles to SP, meeting the


normal at P
in O, and from O draw OQ at right angles to
AP; then the chord of curvature through A is 4:PQ;
drawing AZ perpendicular to TP,
PQ OP : :: AZ :
AP,
or PQ.AP=OP.AZ.
Again, SP PO AZ AT,
: :: :

or OP.AZ=SP.AT;
also PY SP AT TY, : :: :

or PY^=SP,AT=PQ,AP,
Hence 4PQ : 4PF :: PF : ^P.

4. The diameter at P and SP are equally inclined to


the normal at P ; hence chord of curvature parallel to the
axis is equal to 4SP.
Curvature, 69

5. Let QM be the ordinate of Q.


Then QM MG PN NG, : :: :

or PN.MG = 2AS.QM=pn.Mg,
Also Nn = 6r^.
Hence MG :
J/^ pn PN,
:: :

or Jf(3^ :
G^^ ::
pn pn-PN. :

Therefore MG :
pn :: Nn pn-PN : ::
pn^-PN^
: 4:AS (pn-PN) :: pn + PN : 4AS.
But QM MG : :: PiV^ :
2^/$^;

hence QM : 2PiV :: pn (pn + PN) : ieAS\


If P and p approach each other indefinitely, Q is the
centre of curvature,
and PQ PN+QM PG PiV: : :: :

But QM PN PiV2 4AS^ AN AS;


: :: : :: :

hence QM+PN PiV^ ^iV^+^^ AS aSP AS.


: :: : :: :

Hence PQ : PG^ :: SP : AS :: PG^ : SR\


6. If PK be perpendicular to the directrix and S' the
farther focus,

SP : P^:: 04 :
CX\
hence ^P=i^(7,
and SP = %AG.
Therefore PE. PS' = %AC^ = 2SP S'P = 20D \ .

or the circle of curvature passes through aS".

7. If P F be the chord of curvature through the focus


and pp' the focal chord parallel to the tangent at P,

PV.AC=2CD\
or PV. AA' = DD'^=AA\ pp';
hence PF=pp\
8. Let Q be the middle point of the chord and Q3f its
ordinate.
70 Curvature,

IfPNP' be the double ordinate, P'Q is a diameter.

If PQ meet the axis in W^


WM=TN=NW',
hence AM =6 AN,
and =
QM^ PN^ = 4AS.AN=iAS,A3I;
or Q lies on a parabola of which A is vertex and AS axis.
Produce SA to S' making S'A = 3AS, and let PQ meet
the tangent atA in Y\
Then ^ F'^=|PiV2 = M/S'. ^iV=3/S"^ AN=S'A.AW. .

Hence S' Y' is perpendicular to PQ, and PQ envelopes a


parabola of which S' is focus and A vertex.

9. DR
and PCP' are equally inclined to the axis, and
Z>, R, P, P' lie on a circle ;

hence PR, DP' are equally inclined to the axis.

So DQ, PD' are equally inclined to the axis ;

hence PR, DQ are parallel, since DP', PD' are parallel.

10. PG=CD^CP.
Hence the radius of curvature varies as CP^,

11. LP, PG^ PT,Pt = CD\


Hence PL is equal to half the chord of curvature in
direction PC.
CP,CL = CP,PL + OP''=CD'^+CP^=AG' + BC\
12. The circle on PE as diameter touches the curve at
P and goes through Q when Q coincides with P the circle
;

becomes the circle of curvature.

13. If the tangents at two near points P and Q meet


m T,
TP : TQ :: CD : CE,
Curvature. 71

Hence the difference of TP and TQ is very small com-


pared with either, and if a circle be described, of which the
intersection of normals at P and Q is centre, to touch TP
and TQ at P
and Q, when P and Q coincide this becomes
the circle of curvature at P.

14. If (7 be the centre of curvature at the vertex,


AG = 2AS.
If PR be the tangent,
PR^ = CP^-CE' = PN^ + CN^-4AS^
= 2AC.AN+CN'--'AC' = AN'i
15. PGP' and the tangent at P are equally inclined
to the axis.

16. If be the centre of curvature,


PO''--^AG .G^PF.GD.
But PO,PF=GD\
hence PO = GD=PF,
Hence, if with centre G and radius GP, such that
GP^ = AG'' + BG^-AG. BG,
a circle be described, it will meet the curve in points, at which
the radius of curvature has the required value.

17. If PF be the chord and GQ be drawn parallel to


Pt,
2GD^ = PQ.PV=^Gt.PV,
But Gt,PM=^BG\
hence PV PM: :: 2GD^ : BG\
18. If PQ be thecommon chord of the ellipse and
circle of curvature, TPt must make equal angles with both
axes, or GT= Gt,
Make the angle AGP such that
PN NG : :: BG'^ :
AG^)
then GP will meet the ellipse in the point required.
19. SP is one-fourth of the chord of curvature through
Sy hence PQ is half the radius of curvature.
72 Curvature,

20. If Pr be the chord,


PV.PD = 2CD\
But PN=ND,
hence PV : CD :: (7i> : PiV.

21. P(9 : PP PK:: : P(7 :: PP :


Pg,
or PP. PP= PG.Pg = CD' ;

hence P is the centre of curvature.

22. Draw SQ at right angles to SP to meet the normal


in Q ;
let the tangent at P meet the directrix in Z.
Then PL PS : :: ZP ZS PQ PS.
: .: :

Therefore PL = PQ = half the radius of curvature.

23. If CB bisect PQ, PCE is a right angle and


PF.PE=CP'' = CD'']
hence PQ is equal to the diameter of curvature at P.

24. OP : (7P
PF, :: PQ :

or OP.PF=CP''=CD^',
hence is the centre of curvature at P.

25. normal at
If the meet in ^; ^, P BG P, H, K
lie on a circle
hence, vrhen ;
coincides with B, P K becomes
the centre of curvature at P.

26. If HT be produced to H' making iPr= TH,


TQ, PH' will be parallel ;

hence PR RS iP^ T^S' TH TOl : :: : :: :

Therefore PR.PS wTH .ITG ZTP 2(7F:: HP :: : :


2CA,
or 2PR .AC^PS .HP=CD\
Hence PR is one-fourth of the chord of curvature through
S.
Curvature. 73

27. If be the centre of curvature at A,

SO : SA :: SA : AX.
Hence curvature of ellipse is greater than that of parabola,
and curvature of parabola is greater than that of hyper-
bola.

28. By Ex. 6, SP--^ACy


and SP : CY' :: FE : PC.
Hence PB : EP' :: 3 : 1.

29. Let CQ' be conjugate to CF and CP' parallel to


PQ.
Then OJ^^ : OF^ :: OQ.OP : OF' :: CP'' :
CQ'""

:: CZ>2 :
CQ'^ :: TP^ : TF"" :: T^2 .
77^2,

Hence TFOF is cut harmonically.


30. Project the angle between the common diameters
into a right angle ; the ellipses obtained will be inscribed,
symmetrically, in a square, and will therefore be equal.

31. SB^ = Py^ + {EP - Syf


= SP^ + EP'-2EP.Sy
^EP^-S.SP^ Art. 160.

32. Let F be the middle point of PE, the radius of


curvature at P.
Then PF.SY= SP^, Art. 160.

.-. SY SP SP PF,
: :: :

,*.
SYP, SPF are similar triangles, and the angle PSFis
a right angle, so that the locus of S is the circle, diameter
PF.
CHAPTER VIII.

PROJECTIONS.

1. The theorems are obtained by projection from the


following properties of the circle.

Art. (65) Every diameter of a circle is bisected at


the centre, and the tangents at its extremities are parallel.

(70) If the chord of contact of tangents from T to a


circle meet CT in N, GT CN= CA\ .

(71) If M,t correspond to M\ V on the circle,


CM : CM' BC AC
:: : :: Ct : Ct' \

therefore CMXt=BC\
(73) If the chord of contact of tangents from T to
a circle meet CT in V
and CT meet the curve in P,
CT,CV=CP\
(74) A
diameter of a circle bisects all chords parallel to
the tangents at its extremities.

(75) If a diameter of a circle bisects chords parallel to


a second, the second diameter bisects all chords parallel to
the first.

(76) Art. 77 is meant.


If PCP\ BCD' be diameters of a circle at right angles,

QF2 : PV, VP' :: CD'^ : CP\


Projections, 75

(78) If CA, CB be radii at right angles,

CM PN AG BG : ;: : :: GN DM,:

(81) The area of a square circumscribing a circle is


constant ^nd equal to the rectangle contained by diameters
at right angles.

(82) If (7Z>, GP be radii at right angles and the tangent


at P meet a pair of radii at right angles in T and if,

PT,Pt=GL\
(83) Cor. 1. The two tangents TP, TQ from any
point are equal, and the parallel diameters AGA\ BGB'
are equal,
.-. TP : AGA' :: TQ :
BGB;
and these ratios are unaltered by projection.
Cor. 2. In the circle TGT' is a right angle,
and .-. PT,Pr=GP^ = GD'',
GD being parallel to TPT.
(84) Taking ABA' as a semicircle, EGF is a right
angle; project on any plane parallel to the line AGA\
2. If a parallelogram be inscribed in a circle its sides
are at right angles. The greatest rectangle than can be
inscribed in a circle is a square having its area equal
to 2AG^ hence the greatest parallelogram that can be
;

inscribed in an ellipse has its area equal to 2 AG. BG.

3. The theorem is true in the case of a circle, and


follows by projection.

4. The greatest triangle which can be inscribed in


a circle is an equilateral triangle of which G is the centre
of gravity.
Produce PC to
F, making 2GF=PG;
then, if be the ordinate, PQQ! is the greatest tri-
Q VQ'
angle which can be inscribed in the ellipse having its
vertex at P,
76 Projections,

5. If a straight line meet two concentric circles, the


portions intercepted between the curves are equal.
6. The locus of the point of intersection of tangents s^t
the extremities of diameters of a circle at right angles is a
concentric circle.

7. The locus of the middle points of lines joining


the extremities of diameters of a circle at right angles is a
concentric circle.

8. If CP, CD be radii of a circle at right angles and


CA bisect the angle PCD, the tangent at A meets CP in
T such that =2
PD^ A T\
9. If a chord ^Q
of a circle be produced to meet
the diameter at right angles to CA in and be parallelCP
to^e,
Aq.A0^2CP\
10. If to a circle and
OQ, OQ! are tangents be R
a diagonal of the parallelogram of which OQ^ OQ' are
adjacent sides, then if R
be on the circle the locus of is a

concentric circle.

11. If a parallelogram be inscribed in a circle and


from any point on the circle straight lines are drawn
parallel to the sides of the parallelogram, the rectangles
under the segments of these lines made by the sides are
equal to one another.
12. If a square circumscribe a circle and a second
square be formed by joining the points where its diagonals
meet the circle, the area of the inner square is half that
of the outer. And if four circles be inscribed in the spaces
between the outer square and the circle, their centres will
lie on a concentric circle.

13. If a rectangle be inscribed in a circle so that the


diameter bisecting one pair of sides is divided in a constant
ratio, the area is constant.

If a parallelogram circumscribe a circle and one of


14.
its diagonals bear a constant ratio to the diameter it
contains, the area is constant.
Projections, 77

15. PtQ/2 is a triangle inscribed in a circle, the centre


being the intersection of lines joining the angular points to
the middle points of opposite sides. If PC, QC, meet EG
the circle again in P\ Q, R\ the tangents at P\ Q\ R\ will
form a triangle similar to Pi^Ry its area being four times
as great.

16. The locus of the middle points of chords of a


circle passing through a fixed point is a circle of which the
line joining that point to the centre is diameter.

17. The ellipse which touches the middle points of the


sides of a square (i.e. a circle) is greater than any other in-
scribed ellipse.

18. If a polygon circumscribe a circle, its area is a


minimum when any side is parallel to the line joining the
points of contact of adjacent sides.

19. The greatest triangle which can be inscribed


in a circle has one side bisected by a diameter and the
others cut in points of trisection by the diameter at right
angles.

20. ^^ is a given chord of a circle, C any point


of the circle, the locus of the intersection of the straight
lines joining A, B, C
to the middle points of BC, CA, AB
is a circle.

21. If CP, CD are radii of a circle at right angles, the


circle on PjD as diameter will go through (7.

22. The theorem is true in the case of a circle inter-


secting a concentric rectangular hyperbola, and follows

generally by projection.

23. If V is the middle point of Qq, project CVQ into


a right angle.

24. If PT, pt are tangents at the extremities of a


diameter Pp of a circle, then if any diameter meet 2" in P T
and the diameter at right angles meet pt in t, and any tan-
gent meet PT in T and in jt?^ t\
PT PT :
wplf \
pt.
78 Projections.

25. If (7P, CD be radii of a circle at right angles


and P/>, Dd be drawn parallel to any tangent, and any line
through G meet Pp^ Dd and the tangent in p, d and ty

Cp^ + Cd^=Ct\

26. If ACA', BC and CD, CP be pairs of radii of a


circle at right angles and if BP, BD be joined, also
AD, A'Pj the latter intersecting in 0, BDOP is a parallelo-
gram.

27. If TM be perpendicular to SP,


TM TP SY SP BG
: :: : :: :
CD;
hence TP : (72) is constant.

be taken on a tangent to a circle such that


If a point
its distance from the point of contact is constant and
therefore proportional to the parallel radius, its locus is a
concentric circle.
CHAPTER IX.

CONICS lA^ GENERAL.

1. TN XN: SR SX SP XN; :: : :: :

hence SP=TN.
Also TP TP' = 7W^- PN^ = aS'P^ _ PiV^s ^ ^;v^2^
.

2. Draw Pm, Qn perpendicular to the directrix.


Then PR :QN::KP:KQ::Pm:Qn::SP:SQ::PM:QN,
or PR = PM.
3. PS.SQ AS.SA' (7p2
: :: :
C^^^
and PQ.SR^ISP.SQ.
Hence PQ varies as Cp\

4. Let Pj9, Qq intersect in 0,


Then QO.Oq :
PO.Op :: rp2 :
TQ^ :: QO^ : POl
Hence TO bisects pq as well as PQ, and is a diameter
and goes through t

5. Since RS is the exterior bisector of the angle


P'SQ\
SP" : SQ' :: RP' :
/2Q'.
80 Conies in General,

6. Let S'T, ST meet PQ in EE\ and let TF, TG be


the common interior and exterior bisectors of the angles
ETE\ PTQ.
Bisect FP in 0.

Then OE OW = OF^ =OP.OQ.


.

Kow RP EQ PE EQ, : :: :

and RP RQ : :; P^' :
E'Q.
Again, OP
OE OE' OQ; : :: :

hence P^ 0^ E'Q OQ. : :: :

Also OP OE' 0^ OQ : :: :
;

hence P^' OE' ^Q OQ. : :: :

Therefore RP,PR RQ.QR PE.PE' EQ, QE : :: :

:: 0^. 0^' :
OQ^ :: OP^ :
OQK
Hence TF bisects the angle RTR\ and the angles
RTP, RTQ are equal.
7. TS and ^aS' are the interior and exterior bisectors
of the angle PSR,
Hence if ST meet PP' in P,
RK TP ^P EP ^P
: :: : :: : EP' :: ^P' :
PT,
or RK^KR,
8. If PP, D'E' are perpendicular to aS'P, SP',
then SE=SE'.
Hence PP, D'E intersect on the bisector of the angle
PSP\ which is ST.

9. If PP' meet the directrix in K, PP' is har-


monically divided at S and K,
Hence any chord through S is harmonically divided by
the directrix and the tangents at P and P'.
10. Draw /SF perpendicular to the tangent, then since
SO : CY :: SA :
AX,
the locus of (7 is a circle.
Conies in General. 81

11. Let Fp meet the curve in P', and let QP' meet the
directrix in q\
Then since pS, q'S are the exterior bisectors of the
angles PSP\ QSP\ the angle pSq' = \i2i\i PSQ=pSq;
hence «$ and gf coincide.

12. SL : SP FT FP TN P^.
:: : :: :

Therefore aSX : TN aS'P P^ SA ^X


:: : :: :

13. FN' : AC^-CN^ :: <7i?2


.
q^-i
..
(752
.
q^'^

: :
/??i2 : Cn^ - (7a^.

Hence, if FN=pn, AG'^ CN^= Cn^- Ca\


or CN^ + Cn^^CA^-^Ca\
14. Qi? : LG :: PQ : PG^ :: PM : PiV.
Hence QR Pi^ : :: Z(5^ : FN :: /SG^ : SP :: aS'^ : ^X
15. Draw KV parallel to the axis.
Then VK : VP :: GS : /S^P ;: SG' :
aS'Q :: VK :
TQ,
or PV=VQ.
Also PZ : Pil/ :: PK PG PT : :: : PS;
hence 2PZ PS= PM. PQ==SR.PQ = 2SP
. . SQ,
or PL = SQ.
Hence /ST= FZ, and the diagonal of the parallelogram SL
goes through V.

16. If PQi2 be the triangle and S the focus, make the


angles QSr, QSp each equal to the supplement of PSR,
Then, if PSq = PSr, p, q and r are the points of contact.

17. By Ex. 20, Chap, i., if KV be drawn parallel to the


axis, PF=Fe.
Hence FN PL PK PG PV : :: : :: : PS,
Hence 2PiV^. PS= SR,PQ = 2SP .
SQ,
or = SQ. PN
Hence SN=2SV, and the locus of iV is a similar conic.
B. c. s. 6
82 Conies in General.

18. Let CT, CT meet the curve in p, d.


Then CT.PR = 2Cp\
and Cr,QR=2Cd\
Hence the triangle
CTT : 2 triangle Cpd :: 2Cpd PRQ,
:

or the triangle

CTT : AC. EG :: ^(7. ^(7 : the triangle PRQ.

19. Let S be the centre of the circumscribed circle, H


the ortho- centre,

then the feet of the perpendiculars from /S'and H on AB,


BC, CA lie on the nine-point circle,

and the angle /S'^4^ = complement of C—HAC.


Therefore with S and H as foci a conic can be inscribed in
ABC.

20. Let SV meet the directrix in Q and PK'vn Z, let


QP meet the axis in G.
Then PZ SP SG SP SA AX; : :: : :: :

hence PZ PK .S'^' AX^ : :: :

Now CX PZ PK SA" ^X^,


/S'(7 : :: : :: :

or (7 is the centre.

21. DE.DF DG^ : :: ^^2 .


^(72 ..
2)^3^2
.

i>(72,

or DG^= DE.DF.
22. By Ex. 74 on the parabola, if PGQ be the chord of
contact,

DF : i^(? :: PG :
G^Q :: G^P :
FE,
or FG''=FD.FE,

23. If ^^^ be drawn parallel to DTP,


DP'^ :
.^jc»
.
^g :: DF : j&P.
Conies in General. 83

For DF
is parallel to a generating line VM of the cone of
which the hyperbola is a section.
Draw Dim, LEM in the plane VFEM to meet the cone.
Then the sections of the cone by the parallel planes, IPniy
pLQ are similar,
and Dm = EM.
Hence
DP': Ep .Eq Dl.DmiEL, EM Dl EL
:: :: : :: DF : EF.

Again, Ep.Eq EQ' PT" TQ^ EK^


: :: : :: :
Eq^\
hence ^JT^ ^ Ep.Eq if ^jt?^ meet PQ in iT.
And DG" G^2 :
..
j)^2 .
^^2 ..
2)p2 :Ep.Eq::DF: EF.
Therefore FG^ = FD FE. .

24. Let the tangents at P, Q and R meet JS^J5 in p, q


and r.

Then Ep.Eq^ EF\ if PQ meet EB in F;


also EB^=Er Ep, Ea^ = Er Eq. . .

Hence EB^ : EC^ :: Ep :


Eq, a constant ratio.

By Ex. 23. the same proposition is true in the case of


an hyperbola if EB be parallel to an asymptote.

25. GK being perpendicular to SP,


Pk PK Pg PG AC* BC
: :: : :: :
;

.'. Pk is constant.

Also kL : Pk :: SG : SP ;
.'. kL is constant.

Let the fixed line meet the curve in


26. and Q, and P
let the tangent at P
meet SL in D, and the directrix
in F; then, Art. 11, aS'Z) :: /SZ SX. : SF :

The angle PS'P a right angle, so that


is is a fixed SF
line,and, SX being a fixed line, the ratio of iSF to is SX
constant ; SD
.*. is constant and is fixed. D
The envelope of PG is therefore the parabola, of which
D the focus and PQ the tangent at the vertex.
is

6—2
CHAPTER X.

HARMONICS, POLES AND POLARS.

1. If a OB be the common chord and PQOpq any


transversal,
F0,0p = A0,0B = Q0.0q.
2. OA is perpendicular to B'C and meets it in D,
OA OD = square on radius
. of concentric circle = OB OB.

= OC,OF,
Therefore OD = 0E= OF.
3. Draw ABC to be bisected by OB in B, BEG to
AO produced to be bisected by (9(7 in E, and BFK to CO
produced to be bisected by OA in F.

Then any one of the straight lines drawn through O


parallel to AC, BG^ BK will form a harmonic pencil with
OA, OB, OC.
Draw BL, J5i»f parallel to OA, OC to meet OC, OA in L
and M,
Then since OE=EL, OF=FM, EF is parallel to LM
and is therefore bisected by OB and is also parallel to ABC.
Hence the pencil BC, BE, BO, BF is harmonic.
4. Let the meet in P, bisect AG in E.
circles

Then EB.ED = EC = EP%


hence the circles cut at right angles.
Harmonics Poles and Polars.
f 85

6. By Art 182, PQ, AE, BD intersect in A, and A


{B, Ey Q, F} is harmonic.
6. AP, BQ and PB^ AQ meet each pair on the polar
of on which G lies ;

And the pencil formed by OB, CO, CA and the polar of


O is harmonic.
7. If ^, ^ be points of contact and the third conic
meet AB in C and />, A and B are the foci of the involu-
tion, P, Q are conjugate points.
Hence ACBD is a harmonic range.

8. Let the common chords meet in E, and let EPRQ


be a tangent at R then since the common chords are one
;

conic of the system, E and R are foci of the involution EPRQ


and EPRQ is a harmonic range.

9. Let rp, TQ be the tangents, TE any then


line, F
the pole of TE lies on PQ and PEQF is a harmonic
range.

10. If the tangent at P meet the asymptotes in L and


L\PL = PL\
Hence the pencil CD, CL, CP, CL' is harmonic.

11. A
diameter is bisected at the centre: and the
polars of the extremities of a diameter intersect at infinity.

12. If jTbe the pole of QR and H


the second focus of
the conic which touches the ellipse at Q,

PQ + QH=SQ-¥QS''y
or HS'=SP,
Therefore HS' + S'P = S'P + SP ;

or /S" is on the conic of which ^is focus.

Again the angles TS'R, TS'Q are equal, and PS\


S'H are equally inclined to TS' hence TS' is a tangent
;

at/S'.

Therefore T lies on the directrix of the conic of which


P, H are foci.
86 Harmonics, Poles and Polars.

13. Let ST, S'T meet FQ in E, E' \ then the angles


JSrrPjj&'T'Q are equal.

Now the ranges RPEQ, R'QEP are harmonic and


QTP common to the two
is pencils; hence the angles
i2'rQ,i2rP are equal.
14. The middle points of all chords of the cone parallel
to the given line, lie in a plane through the vertex, let this
plane meet the given line in P
and any section through it
in A and A ',

Then the pole of the given line lies in AA'P and


Q
AQA'P a harmonic range. Since VA, VA' are fixed
is

generating lines, VQ is a fixed straight line.

15. If Tpq be the chord, P its pole, then PN the ordi-


nate of P is the polar of T.

Let GP meet pq in v and the curve in Q,

Let QM, QG' be the ordinate and normal at Q,

If PG be drawn perpendicular to pq, it is parallel to


QG';
hence GG GG' GP GQ GN GM;
: :: : :: :

Therefore GG GN GG' GM SG"" -^ICl


: :: : :: :

Hence G is a fixed point.


16. Let the polar of Q meet the conjugate GFD in B.
Draw QQ' parallel to GP.
Then P^ PQ : :: EF PP; :

and PG PQ : :: (7P :
PF;
hence ^G^ PQ: :: GE :
PF;
or EG.PF=PQ,GB.
Now Q is on the polar of ^, since 22 is on the polar of Q,
Hence PQ,GB=GQ\ GB = GD\
Hence EG is equal to the radius of curvature at P.
Harmonics, Poles and Polars. 87

17. If be the orthocentre of ABC and A'B'C the


reciprocal triangle, B'C\ C'A', are perpendicular to A'B
OA, OB, OC
respectively, and BG, CA, are perpen- AB
dicular to OA', OB', respectively. OC
Hence ABC and A'BC have their sides parallel and O
is the orthocentre of each.

18. If pPSQq be the focal chord and the tangents at


P and Q meet in T, TS is perpendicular to PQ, hence the
tangents at p and q meet in T.

19. This theorem is the reciprocal of the following :

if two they have two common tangents if


circles intersect :

one circle lie entirely within the other, they have no


common tangents. Reciprocate with respect to a point on
one circle and within the other.
20. If /, C be the centres of the inscribed and circum-
scribed circles, and CI meet them in r, r and B,
'
re- B
spectively, then if AA' be the major axis of the ellipse into
which the circumscribed circle is reciprocated,
IA.IB = Ir\ IA'.IB'=Ir\
Cr'=CB^-^CB,Ir.
Hence lA : Ir :: Ir : CB-CI :: CB + CI \ ICB,
and lA' : Ir :: Ir : CB^CI :: CB-CI ; 2CB.
Hence A A' : Ir 2CB 2CB
:: ;
;

or AA' = Ir.
21. The four circles which circumscribe the triangles
of a complete quadrilateral meet in a point.

22. See Ex. 30 on the parabola, or by reciprocation.

23. See Ex. 46 on the ellipse, or by reciprocation.

24. Let CPP\ COO' be perpendicular to the polars


of P and O.
Draw OX, OA perpendicular to CP and the polar of P ;

P Y, PB perpendicular to CO and the polar of O.


Then CO' : CP' CP CO CY CX.
:: : :: :

Hence CO'-CY CP'-CX CY CX CP


: :: : :: : CO
or FB OA CP CO, : :: :
88 Harmonics, Poles and Polars,

25. See Ex. 18 on Chapter I.

26.(1) If a quadrilateral circumscribe a conic a pair


of opposite sides subtend at the focus angles which are
together equal to two right angles.
(2) If we reciprocate with respect to the focus S the
new theorem is, if Q be taken on a circle and QL be drawn
such that the angle SQL is constant, QL envelopes a conic
of which S is focus.

27. The envelope of chords of a circle which subtend a


constant angle at a fixed point on the circle is a smaller
concentric circle.

28. Two circles, such that a point can lie within both
cannot have more than two common tangents.
But if the circles be such that all points lie without both,
or within one and without the other they may have four
common tangents.

29. If a straight line meet the sides of the triangle


A'B'C in X, M, N the circles circumscribing the triangles
A'BC, A'NM, B'NL, G'LM meet in a point.

30. If points P', Q' be taken on a circle of which C is


the centre, P'G will meet the line drawn through Q^ at right
angles to P'Q' and Q'G will meet the line drawn through
F' at right angles to P'Q' on the circle.

31. If S be the orthocentre of the triangle and ABC


circlesbe described with centres A and B
passing through
C, S will lie on the radical axis of the two circles.
If we reciprocate with respect to S we see that if with
the orthocentre of a triangle as focus we describe two
conies each touching a side of the triangle and having the
other two sides as directrices, the conies will have a parallel
pair of common tangents and therefore their minor axes
equal.

32. If a system of circles have two points in common


the locus of their centres is a fixed straight line, and the
polar of a fixed point meets the radical axis in a fixed
point.
HarmonicSy Poles and Polars. 89

If the tangent and normal at


33. meet P QR in T and
6?,the range TRGQ
is harmonic, since TF, PG bisect
the angle QPR.

Hence PG is the polar of T,

Hence the pole of QR lies on PG since the pole of PG lies


omQR.

34. If the tangents at P and Q meet in T and TA


meet P^ in Z, the range DPLQ is harmonic ;
hence the
pencil TD, TP, TL, TQ and the range DBAO are har-
monic.

Therefore ABBC is a harmonic range.


35. If the pencil joining BPA
Q to any point on the
curve is harmonic, the pencil formed by joining them to
any other point on the conic is harmonic.
For if BK, PK^ AK^ QK
meet the directrix in hpaq,
hpaq is a harmonic range, provided KEBPAQ
be a har-
monic pencil.
And the angles hSp, pSa, aSq, qSb are half the angles
BSP, PSA, ASQ, QSB ;

Hence the pencil Sb, Sp^ Sa, Sq is the same wherever


A" be taken on the curve.
Now PQ goes through the pole of AB : let PQ meet
AB in R.
Then if The the pole of PQ, TARB is a harmonic range.
Therefore the pencil joining Q BPAQ is harmonic
to ;

hence the pencil joining q to BPAQ harmonic. is

Hence Pq bisects AB since AB, qQ are parallel


36. Four circles can be described so as to touch the
sides of a triangle, and the reciprocal of the radius of the
inscribed circle is equal to the sum of the reciprocals of the
radii of the other three.

If the triangle be equilateral the inscribed circle touches


the three escribed circles.
90 Harmonics, Poles and Polars.

37. If the tangents at P and Q meet the axes in T


and Vf the angle
PSQ = SQ V- SPT= SVQ- STP ^ VST.
li SW ho perpendicular to PQ\ the tangents at the
vertices intersect in W.
Draw aS'FZ perpendicular to the tangents at P and Q.
Then WSP\ WSQ' are supplementary to WYP\ WZQ,
Hence P'SQ', PSQ are supplementary.
If two P, Q the angle between the
circles intersect in

tangent at P, Q is equal to the angles which the centres


subtend at S and supplementary to the angle which PQ
subtends at the other point of intersection.

38 and 39. If from any point P


in the radical axis
tangents be drawn to the circles, and a circle be described,
with centre P
and radius equal to the tangent, this circle
will intersect the line of centres in two points JE and F
which are the limiting points of the system.
Take A
at centre of one of the circles, and at the M
point where the radical axis intersects the line of centres.

Then, P U and P U' being the tangents from P to the


circle
PM^ + ME^ = PE'^ = PU^ = PA^-AU^',
.-. ME^=AM''-AU'^MF';
,\ AE.AF=AM^-EM^ = AU\
.'. the polar of P passes through E,
Reciprocating with regard to F, the pole of uU\ i.e. of
the fixed line through E, is the centre, which is therefore
fixed, and the conies are confocal.

Therefore, if we reciprocate with regard to either limiting


point we obtain confocal conies.

40. If perpendiculars be drawn from A, B, C to BC,


CAf AB these lines will meet in a point 0, and the circles

circumscribing ABC, OBG, OCA, GAB are equal.


41. If the tangents at P and Q, points on a circle
intersect at a constant angle, and lines be drawn through
Harmonics, Poles and Polars, 91

P aiid Q making constant angles with the tangents at P


and Q respectively, this pair of straight lines will intersect
on a concentric circle.

42. If two circles intersect in A and B and PQ be a


common tangent and QB, PA meet the circles in C and 2>,
then PC, QD are parallel.

43. If from any point on a circle circumscribing a


triangle perpendiculars be drawn to the sides of the tri-
angle, the feet of these perpendiculars lie on a straight
line.

44. Since the orthocentre is on the hyperbola, DEF


is a self-conjugate triangle and the pole of lies on BC,EF
Hence the pole of BC lies on EF,

45. If AB, CD meet in the fixed point E, CA and BD


in F, and BC and AD in G, then FG is the polar of E,
Hence the centre of the circle lies in a straight line
through E
perpendicular to FG
the polar of with respect E
to both curves.

46. The radius of an escribed circle of an equilateral


triangle is | the radius of the circumscribed circle, and if
SE be the tangent from the centre of the circumscribed
circle to the escribed circle whose centre is D ;

SD = DE+iDE= ^DE.
The proposition in the question is obtained by recipro-
cating with respect to the circumscribed circle.

47. If AD be drawn parallel to the axis to meet BC,


AD is bisected at D^ where it meets the curve.
'

Hence the tangent at Z>' is parallel to BC and bisects


AB and AC
Since a straight line intersects a conic in two points and
two tangents can be drawn from a point, the reciprocal
polar of a conic with respect to another conic is a third
conic.
92 HarmonicSy Poles and Polars,

Now by Ex. 44, if a rectangular hyperbola circumscribe


a triangle DEF it will go through the ortho-centre and
ABC the triangle formed by joining the feet of the perpen-
diculars is a self- conjugate triangle, and O is the centre of
the circle inscribed inABG. If we reciprocate with respect
to O the reciprocal conic is a parabola, since it has one
tangent at an infinite distance and ABC
is a self-conjugate

triangle.

The tangents from are at right angles, since the


hyperbola was rectangular, hence is on the directrix.

The locus of the poles of the lines at an infinite distance,


that is, of the centres of the hyperbolas, was the circle cir-
cumscribing ABC.
of O with respect to the
Hence the envelope of the polars
parabolas is an ellipse inscribed in ABG
having O for a
focus. Since is now the centre of the circle circum-

scribing ABCy the auxiliary circle of the eUipse is the nine


point circle.
MISCELLANEOUS PROBLEMS.

1. If andS H
be the rifle and target, and the P
hearer, the difference of the times in which sound travels
from iS' and to ^ P
is equal to the time of the bullet's

transit from S to II.


Hence HP-SP is constant, and the locus is a hyper-
bola of which S and // are the foci.

2. Let tp^ tq be the tangents parallel to and P'Q PQ


and let qt meet inr the diameter through p) then qt^tr,
and
..
Pi22 .
PQ2 tr^ .

fp2 :: tq^ :
tp^
:; SP'.SQ' :
SP.SQ
:: P'Q' :
PQ ; Art. 17.

.\PR^=PQ.P'Q\
3. QN CM BC AG DM
: :: : :: CN,
:

hence QN+DM NM BG : :: : AG,

4. If GVBD be conjugate to PQ,


PQ,Qp=PV^-QV\
Hence PQ.Qp CD^-CE^ : :: Ci?^ :
(72)2,

GR being parallel to PQ.


5. CiV^ : iVX :: .S'^ : AX :: /S'i? : SX.
Hence Q lies on the tangent at the extremity of the latus
rectum.

6. PN^ : AN. NA' :: BG"- : -4(72

and QN.PN=AN.NA\
Therefore QN'' : AN.NA' :: AG^ : BG\
94 Miscellaneous Problems.

7. Let AP, QB meet in i?, and draw RV parallel


to POQ.
Then RV VA PO
: :: :
AO,
and RV VB=QO: : OB.
Hence RV'^^VA.VB,
since AO.OB^PO.OQ.
Therefore QR lies on a concentric rectangular hyperbola.

8. The line joining T to the intersection of the


normals at P and P' bisects PP' and therefore passes
through the centre.

9. If the tangent at P meet the tangents at A and A'


in Tand T and TS, T'S' meet in Q,
the angle SS'Q = TS'A'= T'S'P ;

QSS'^AST=TSP',
and SPT=jS'Pr.
Hence S, P, S' are the feet of the perpendiculars of the
triangle TQT.
Therefore QP is perpendicular to TP.
10. Make the angle PSF a right angle, then the
tangent at P
meets the directrix in i^: if a circle be
described with centre P
and radius PK
such that the ratio
SP : PK
is equal to the eccentricity the directrix is a

tangent from F to this circle.


Two tangents can in general be drawn.
If the angle SPF
be such that :: SP
SA AX^ : PF :

only one conic can be constructed there are two positions


;

of PFequally inclined to SP
corresponding to this case.
If the eccentricity be unity, one conic is a line parabola
through S.

11. If PK be drawn perpendicular to the directrix of


the parabola SP = PK,
hence HM=^HP + PJr= AA\
Miscellaneous Problems. 95

Tlierefore the directrix touches a circle of which H is

centre.

12. Draw UN perpendicular to the minor axis.


Then
CN.AC=SR.AG=BC^=AC^-S(P^AC^-RN^;
Hence BN'=AC{AC-CN},
or R lies on a parabola.
13. If ST, FQ meet in H, HTRS is a harmonic
range.
But OH, OT,OV, OS is a harmonic pencil.
Hence V passes through R.

14. Let ACA' be the diameter bisecting the parallel


chords QNy etc. in iV, etc.

Then PN^ varies as QN\ that is as AN. NA\


Hence the locus of P is an ellipse. The locus will
be a circle if PN=QN, that is if the vertical angle
is a right angle.

15. If PP\ QQ' be the double ordinates of the


given points, P, P\ Q, Q' are fixed points, and since
the ellipses are similar, the corresponding points of
the auxiliary circle, at which the major axis subtends
a right angle, are likewise fixed points.

16. The ordinates of the point and of the end of one of


the radii are in the ratio of the radii.

17. If P be the centre of the circle and P^ perpen-


dicular to the fixed straight line, the ratio SP PK
: is
constant.

18. Let pqr be a triangle touching the parabola in

Parabolic area PQR = | triangle PqR.


.-.
Triangle PQR^%{PqR- PrQ-QpR),
3PQR-^2{pqr^-PQR);
.*. PQR = 2pqr.
96 Miscellaneous Problems.

19. If a rectangular hyperbola circumscribe a tri-


angle the orthocentre is on the curve, if we reciprocate
with respect to the orthocentre we have the case of a
parabola inscribed in a triangle the tangents from the
orthocentre being at right angles. See Ex. 47, Ch. 10.

20. Produce HA to K making AK equal to AH, then


PK and AL are parallel.

Hence SQ : aSP :: SA : SK :: SA : SA + AJI,


Therefore the locus of Q is a similar ellipse of which
S is focus.

21. If KLMN be the quadrilateral and k, /, m, n the


points of contact, KM will bisect nky Im : and LN will
bisect kl, mn.

Hence klmn, and therefore KLMN, is a parallelogram.

22. The locus of the second focus is a circle of which


the radius = AA- SP.
The locus of the centre which bisects SH is similar, that
is, a circle.

23. Since LC, LL' are tangents, the angles HLC and
SLL are equal.
Again CL . CL' = GH\
hence the angle CHL = GL'H^ SL'L.
Therefore CL : HL :: SL :
LL\
the triangles CLH^ SLL being similar.
24. If the theorem be true in the case of a circle, it will
follow by orthogonal projection for any ellipse.

If PQbe the chord of contact of tangents drawn to


a circle from a point on a concentric circle, the angles
PAQ, PA'Q will be constant, A, A' being extremities
of a fixed diameter.

Let APy A'Q meet in R, and A'P and AQ'mR.


The angle AEA' = APA'-PA'Q,
and the angle ABA'= APA' + PA Q.
Miscellaneous Problems, 97

Hence the loci of R and R' are circles passing through


A and A'.

25. Circumscribe circles to two of the triangles formed


by the intersections of the tangents, these circles intersect
in the focus JS: the pedal line of S is the tangent at the
vertex.

A parabola can be drawn to touch five straight lines, if


the circles circumscribing the triangles formed as above all
meet in the same point S.
26. PF is the same for both curves, and therefore CD
is also the same.

27. Prove that SY. S'Y' is constant.

28. By reciprocation.

29. P, Qj R, R' lie on a circle of which PQ is diameter


and PQ, LL' are equally inclined to the axis. If jt?, p' are
the vertices of the diameters bisecting PQ, RR' in V
and V\ pp' is a double ordinate.
Let VV which is parallel to the normal at /?' meet the
axis in 0.

Let VM, V'M' be the ordinates of Fand V\


'l]ie\iLL'=L'M' + M'0 + MO-LM=20M-=^2ng = 4AS,

30. The bisectors are tangent and normal to a con-


focal conic.

Hence CG,CT=CS\
Reciprocate the following theorem li S, A, B, C
31. :

be points on a circle and with centres A, B, Cand radii

AS, BS, CS circles are described, they will intersect two


by two in points which lie in a straight line.

32. OE EG = SP SG = Sp
: : :
Sg = OE :
Eg.

33. If an ellipse be reciprocated with respect to its


centre, the reciprocal is a similar ellipse having its major
axis in the minor axis of the original ellipse.

B. c. s. 7
98 Miscellaneous Problems.

If we
reciprocate an ellipse circumscribing a triangle
and having its centre at the orthocentre with respect
to that orthocentre, the reciprocal is a similar eliipse,
inscribed in a triangle having its sides parallel to those of
the original triangle, the homologous axes being at right
angles, and having its centre at the orthocentre.

This reciprocal ellipse is similar and similarly situated


to the ellipse inscribed in the original triangle having
its centre at the orthocentre.

34. Q lies on the common circle of curvature,

hence PQ = 4PT.
35. ABf BG are equally inclined to the axis, hence
since the angles at A and G are equal, are equally AD,DG
inclined to the axis.

Hence the tangent at D and A G are equally inclined to


the axis.

Therefore the tangents at B and D are parallel.


36. The volume cut off varies as the area VAA' and
BB' and the area VAA' varies
; as ^4 F. VA' or AD A'D\ .

that is BG\
37. SQ :
Pg :: St :
tg :: SY SP : :: BG GD :

:: PF : AG.
Hence SQ.AG=PF.Pg=AG^,
or AG^SQ.
Let QL be the ordinate of Q and let MQ meet the major
axis in V.

Then GV PN : :: GV GM : :: GL :
GM-QL
:: GL :
PN-QL,
and SP-AG GN : :: SG AG, :

or AG, SP = AG''-rGN. SG=BG''+ GS. SN.


Again GN-GL SP-SQ SN : :: :
/S'/*,

and SP-SQ (7iV 6'(7 AG] : :: :


Miscellaneous Problems. 9^

hence CN-CL CN SN.SG SP.AC;


: :: :

therefore CL CN :: BC^ SP.AC,


: :

or CL SP-SQ BC^ aS^P.^S'^


: :: :

Now SP-SQ PN-QL ^P PiV;


: :: :

hence (7Z :
PN-QL i?(72 PN.SC, :: :

Hence CV.SC=BC\
or F is a fixed point.

38. If /SX, aS'JI/, SN


be drawn perpendicular to
the given tangents, the circle circumscribing LMN is

the auxiliary circle of the conic.

39. If S be the centre of the circumscribed circle, H


the orthocentre, the centre of the nine-point circle bisects
SH\ and if PQR
be the triangle, the angles SPQ, HPR
are equal hence S and // are the foci.
:

40. If Pg, P'g' be the normals at P and P' and


gL^ g'L' be drawn perpendicular to PP\
then PL = RL\ by Ex. 27, Chapter I. ;

hence gG=Gg',
Therefore
2SG : SP + SP' ::
Sg + Sg' : SP + SP' :: SA : AX.
41. Reciprocate the following with respect to S:
ASB a diameter of circle meeting a concentric circle
is

in >S', the opposite sides of the quadrilateral formed by

tangents through and A B


to the inner circle are parallel,
and the tangents to the outer circle at the points where it
meets the tangent at S are respectively parallel to them.

42. If P, Q be two points on a rod and PS, are at QS


right angles to the directions of motion of and Q, then if P
R be any point on the rod the direction of motion of is R
at right angles to SR.
Hence the directions of motion of all points on the rod
envelope a parabola of which aS' is focus and the rod tangent
at the vertex.

1—2
100 Miscellaneous Problems.

43. SP and HQ are parallel to CT.


Hence PaS^ = complement of /S'Pp = complement of CTP.
QHq = complement of iZQg = complement of CTQ.
Hence PSp and QJTg are together equal to the supplement
oiPTQ.
44. Let ST, S'T meet CP in ^? and p\ and (72) in
d and c?'.

Since the angle PTS=d'TD,


and TPp = TDd\
the angles /S'j^C, (7J'/S" are equal, and c?, d', p and j9' lie on
a circle.

45. If the asymptote and directrix meet in 2), SDG is


a right angle and if DP
be the tangent PSD is a right
angle.
Therefore SP is parallel to the asymptote.
46. If PTQ, ptq be two consecutive positions and
TV, ty be drawn perpendicular to pt, TQ respectively,

tV=Pp+pt-PT=PT+TQ + Qq-tq-PT
= TQ + Qq^tq=Ty,
Hence the tangent at Tis equally inclined to PT'and TQ.
Hence T lies on a confocal ellipse.

47. CP and CQ are at right angles ;


hence G, P, 2), Q
lie on a circle.

48. If PF
be the chord through the centre, and
pp' the parallel focal chord,
PV.CP = 2CD\
and pp\GA = 2GD\
Hence PV pp': CA GP.
:: :

49. SP and QH are both parallel to GT ;

hence the angle


pGq =pGT+ TGq = PHQ + QSP.
50. The angle SP F=half the supplement of SPS'=
hsiUPSr.
Miscellamous Problems. ,

, ; ; J.Ol

61.QSP, Q'SP' are right angles and the perpen-


diculars from Q, Q' on SP, JSP^ are equal to the perpen-
diculars on PP\
Hence QP, Q'P' are tangents at Q and Q,' to the parabola
of which /Sis focus and PP' directrix.
And the diameter parallel to PP' is tangent at the vertex,
since it bisects SH. BB' is a tangent since SCB is a right
angle.

52. 8E will evidently envelope a conic of which S


is focus and the given circle auxiliary circle.

53. Join SP, the bisector of bisects SP in V\ POS


since the locus of P
is a circle, the locus of F is a circle.

Hence VO envelopes a conic of which /S is a focus.

54. The angles RSP and QHV are the complements


ofP/S'TandQlTT:
Hence TJaS'P + C^r= supplement of half PSQ^PHQ
= half the angle between the tangents at P and Q, by
Ex. 23 on the ellipse.

55. TS, ZS are the interior and exterior bisectors


of the angle QSR,
and if TQ meet PZ in P, FS is the exterior bisector of
the angle QST.
Hence Q, T, R
lie on a conic of which S is focus and
PZ directrix and ZT
is the tangent at T since ZST
is a right angle.

56. If OE be the radius of the sphere,


OE.AC=2irQ2iOAA'
and varies as OA OA' ov AD
. . A'D' that is as BC\
Hence the iatera recta of all the sections are the same.

57. Let PQ\ QP' meet the ellipse in ZZand F,


then since TP^ = TQ TQ' .

and TQ^=TP,TP\
TP : TP' :: TQ' ;
TQ,
or PQ', P'Q are parallel.
102 Miscellaneous problems.

Hence if OF be parallel to P Q and PQ',


P'V.P'Q : P'P'' :: CF^ :
CD\
and Q'U.qP :
Q^Q-^ :: (7i^^ : GEK
JS^ow P'P^ :
qq^" :: TP^ :
T^Q'^
..
^j^q
.

7^^/^

and CE'^ : (7/>2 :: TQ^ : TP'^ :: TQ :


TQ'.
Hence P T. P'Q .qU.Q'P.: TQ^ : TQ 2 ;: p'Q2
.

pqi^
Therefore PV P'Q Q'U : y. :
QP.
58. If the tangents at P and Q meet in T, CT which
bisects PQ is parallelto P'Q, P' being the other extremity
of the diameter PCP\
Hence the angle PCT= PP'Q = TPQ.
59. /S', P, aS", g lie on a circle,

hence Sg :
P^ :: S'G : aS'P :: aS'^ : AX.
60. ^(7 is parallel to the polar of A, hence AD is parallel
to the axis.
Also the angles SAC, DAB are equal : and the angles
ABDj A SO are likewise equal.

Therefore AC AS AD AB. : :: :

61. RK QN KC NC : :: :

and PN RK : :: iV^(? : KG.


Hence ^C AC : :: KCNG :
NC.KG,
or ^(7 AC ^(7. : ^(9 :: :

Therefore (7i2 :
CQ (7^ CN AC+BC: AC,
:: : ::

or CR=^ AC+BC
Hence if iVP meet /?// in iV^', PiV'^ QiV.
Hence KL passes through P.
Also PL = QC=AC,
and KP = QR^BC
62. CT.CN^CA^
and CT',PN=BC'^)
Miscellaneous Problems, 103

hence CT.CT : CA.CB :: CA.CB : CN.PN.


Hence the triangle CTT' varies inversely as the tri-

angle FCN,
63. By Ex. 6. Chap. VII.
2SP=3ACy
aiid if CB be drawn parallel to the tangent at P,
P£=AC.
64. Let CT which bisects PP' in V, meet the ellipse
in Qy and let Cj& be conjugate to CQ.
Then PV^' : (7F. F^ :: GE^ :
CQ^ :: PF2 :
CQ^-CV\
Hence ce2=(7r. rr+(7F2=6^r. cr,
or 2LP, T'P' are tangents.

65. See Ex. 82 on the hyperbola.

66. If we reciprocate with respect to a focus the


theorem that tangents to an ellipse at right angles intersect
on a fixed circle, we find that if the sides of a quadrilateral
A BCD subtend each a right angle at a fixed point S the
sides envelope an ellipse of which /S'is a focus. If be the
centre,
the angle0^5 = complement of half -4 05 = complement
oiSGB = CBS=SAD.
Hence is the other focus.

67. If A', B, C, D' be the points of contact and


E\ F\ G' the points of intersection of A'G', B' D'\
A'D\B'G'\ A'B',C'D\
then E'F'G' is a self-conjugate triangle.
If BA'A, G'DFmQQt in F
the pole of A'G\ will lie F
on F'G' the polar of E\ since E' lies on A'G' the
polar of F.
Similarly AD'D, BB'G meet in G on F'G' and AC,
BD in E\
Let A'D\ CC'D meet in a, and B'G', ADD in /S.

Then if be intersection of AC, GD\ a8 is the


polar of 0.
104 Miscellaneous Problems,

Now since GB. GD' are tangents and the tangent at C^


meets them, the range CC'DF is harmonic and therefore

the pencil GB\ GC\ G^, GF\

Hence B'C ^F' is a harmonic range.

So AD' aH is a harmonic range.


Hence ajS passes through G'.
Since G' hes on the polar 0, lies on BD, the polar of G.
Similarly AB\ CA' intersect on BD,
68. By Art. 138 the four points in which two rect-
angular hyperbolas intersect are such that any one of them
is the orthocentre of the triangle formed by the other
three hence any conic through the four points is a rect-
:

angular hyperbola.

69. If KVt be drawn parallel to the axis to meet,


PQj ST in F and ^, and if KL be perpendicular to PQ,
PL = SQ and Pr= VQ, hence tV= VK. Therefore ST
SP, SK and the axis form a harmonic pencil.
70. Let DE meet PF in K) and let PD, P'E meet

Then GH is the polar of K^ but K lies on DE the polar


ofi^.
Hence F lies on GH, and FG is parallel to the chords
bisected by PP'.

71. Let RR the common tangent be bisected by PQ


the common chord in 0.

Then RO" = OR'^ =OQ,OP\


hence RR\ PQ are equally inclined to the axis.
Hence PR is a diameter, and the diameter of curvature
= 2PF=2CD,
Therefore CD'- = CD PF= A (7. BC.
.

Reciprocate with respect to S the following theorem


72. :

S is taken on the outer of two concentric circles S Y, SZ ;

are drawn perpendicular to a pair of parallel tangents to


the two circles ; YZ is constant.
Miscellaneous Problems, 105

73. Let the tangents at P and Q meet in T.

Then the angle P'/e(2' = PJ'Q- supplement of P(7Q, by


Ex. 68.
Hence P', C, Q' and li lie on a circle.

74. TN is half the difference of QM and Q'M' and


i?i? is half their sum.

Hence R'P = | Q'M' = ^J?.


Now PN :
Q'M' :: TK : 2A'^/2 :: Pil/ :
QM :: QJf 4SP, :

Hence PiV^ :
Qi»f :: Q'M' : 4>S'P :: PM' :
Q'M'.
Therefore PN^ : PM'^ :: Qil/^ :
Q'3f^ :: PJ/ : Pil/;
or PN''=PM.PM',
75. Let i2i2' F be the diameter bisecting PQ.
Then if PQ meet a common tangent /?p' in O,

Oi?^ : OP. OQ ::
Sp : .ST? ::
S'p' : S'R' ::
0^^ : OP, OQ,
or Op = Op',
76. Let 0, (7 be the centres of the hyperbola and ellipse
to ibCh'f the tangent at (7, then since , Ct-=r.Clr and PN
CN CO = Crt^, the area of the ellipse will be a maximum,
.

when CN,PN is maximum, that is, when CN,NO is a


maximum, or ON=NC,
Hence PP' is a tangent to a similar hyperbola.

77. RP .RP'=Rm-PN^=RN^-4.AS .AN.


Now RN'^ : ^^aS'.^^' :: AN^ AA'\ :

or i2iV^2
.
4^^. jijsf :: jijsf aA\
:

Hence RN'^-PN^ : 4^AS'.^iV^ :: ^W : ^^'.


Therefore RP.RP' : AN.A'N :: 4:AS : A A'.
78. Let the tangent at P meet the confocal conic in Q.
Draw GEF parallel to PQ meeting the normal at P in F,
Then OP PP= (72>2 = ^s^p p^/ ^ ^^2^
.

(7^? being conjugate to CP.


Hence is the pole of PQ with respect to the confocaL
106 Miscellaneous Problems.

79. Let the tangents naeet in U, SU meeting the curve


in Q, and let the tangent at Q meet BR' in T, Then, V
being point of contact of RB\
TSR=TSQ+ l/SP-RSP=TSV+ USP'-RSV
= USP'-RST,
.-.
2TSR=USP'=RSR',
:. locus of T is tangent at Q.
Or by reciprocation of the theorem,
If ABC be a triangle inscribed in a circle, and DE the
diameter perpendicular to AG, DB and EB bisect the
angle B and its supplement
80. Reciprocate with respect to any point S the
theorem that if two points on a circle be given, the pole of

PQ with respect to that circle lies on the line bisecting PQ


at right angles.

81. PQ.PR = PE.PF=AG,BG


and QR = GE^AG-BG,
Hence PQ=BG 2ind PR= AG.
Also ER parallel to GQG.
is

Hence PG GD PG PE BG AG
: :: : :: :

and PG.PF=BGi
Hence GQ, GR are the axes.

82. This is a particular case of Art. 195, since the


second point where AE
meets the curve is at an infinite
distance, hence AE=EK.

83. The circle of curvature is greatest at the ex-


tremity of the minor axis.
Hence BO the direction of the minor axis is given.
And BG.BO = AG'^=SB\ being the centre of curvature.
Hence S lies on the circle of which BO is diameter.

84. Ga : Gh :: Ba.Ac : bA cB .

and Oa : Gb' ;: Ba.Acf : h'A .


c'B,

by Todhunter's Euclid, Art. 59.


Miscellaneous Problems. 107

And Ac . Ac' : Ab.Ah' in ratio of squares on parallel


diameters.
Hence BG is the tangent at a,
This depends on the fact that any chord is bisected
85.
by the diameter through tlie intersection of the tangents at
the-ends of the chord.

86. Let P, Q be the points of contact of parallel tan-


gents to the conic and circle.
Then by Art. 132, the angles PGA, QGA are equal.

87. Let GL, GL be the fixed straight lines, /S'the fixed


point.
Then the angle LSL'^GSL' + GL'S,
hence the angles GL'S, GSL are equal.
Therefore GL : GS :: GS GL\:

or LU touches the hyperbola of which CX, GL' are asymp-


totes and S focus.
88.Since the semi vertical angles are complementary
they touch one another along their common generating
line.

Now EA : AX GE EG :: : :: OE' G'E'


: :: EA : AS'.
Hence S' coincides with X, and similarly JS with X'.

89. Draw GR F perpendicular to the tangent at P, GE


PP' and PN parallel to GE.
bisecting
Then QO.OQ' PO.OP' GL>^ :GR^ :: : :: PO.PF:
GN. GV PO PE "IPO PP'.
V. \ V. :

90. A circle can bo described with centre T to touch


SP, SQ, HP, BQ.
Hence SN-NII=SM-MH,
and TM^ TN bisect the angles at M, N.
Hence TM, TN touch a confocal conic passing through M
andiV;
91. If aS' and H
are the given points, the locus of is P
the conic in which the given plane through aS' intersects the
108 Miscellaneous Problems,

surface generated by the revolution about SH of a conic


of which and S
are foci. H
If ST be drawn perpendicular to the plane to meet the
directrix plane corresponding to yS'in Ty the cone formed by
joining H
to all points of the locus of is a right circular P
cone of which TJri is axis.

92. PG.Pq = BG^ = PQ^;


,'. PGQ and PgR are similar triangles.
93. If OL be the ordinate of 0,
LG GN OL P'N CL CN,
: :: : :: :

or LG LG CN NG AG BGK
: :: : :: :

Hence CL (76^ : :: ^(7^ : AG' + BGK


Therefore GO GP' : :: (7^ : CiV^ AG^^BG^ AG^+BG\
:: :

94. The polygons FaSP and Z'HPZ are similar and F


the perpendicular from G on FZ bisects VZ ;
hence if FLE^be taken on FF' such that VE=ZZ\
then GE=GV'=GZ\
Hence W ,ZZ= W . VB= VG^- rG^ = GA'-GA'\
95. The centre is the middle point of GP,
96. TSQ and
T'/S^'Q are right angles ;

.*. the middle point of TQ is the centre of the circle TSQS'


and is equidistant from S and S'.

97. 7^(2 : TP :: aS'Q :


/S'r,
and TP : TQ' :: ST :
/SQ' ;

.-. TQ,TP TP,Tq : :: aS^CaS^^ : SQ! , ST


\\SQ,Pr '.sq:,pt.
98. Draw NE perpendicular to NM^ and prove that E
is a fixed point in the axis.

99. P and Q are equidistant from the plane of the


circular section of the cone, which contains the centre of
the section.

100. Produce OG to E so that GE= OG ;

then PE is parallel to GZ and EP Y= GZO = OPY; that


is, the tangent to the curve bisects the angle OPE,
Miscellaneous Problems. 109

101. If PEQ be one of the tangents, and ERV the


chord,
EP^ : ER.EF :: EQ^ :
ER.EV,
for each ratio is that of the parallel focal chords.

102. If i^^, F(? be the tangents, i^(ra:QGf) is harmonic,


and EFG is a right angle.
103. Reciprocate with regard to C the theorem, that,
if a circle centre C intersect another circle at right angles
at the point E, and CPQ be any chord, CE^ = CP, CQ.

104. Reciprocate the conies into two intersecting


circles.

105. Reciprocates into the theorem of the existence of


the director circle.

106. If PEQ
be the chord required, and P'EQ' a
consecutive chord, the areas PEP\
QEQ' are ultimately-
equal, and Ey which is the centre of curvature, is the middle
point of PQ.
PQ is therefore the diameter of curvature and is in-
clined to the axis at the same angle as the tangent, i.e. half
a right angle.

107. The pole F


of the straight line is fixed, and P,
the point of contact of a tangent, is the foot of the perpen-
dicular from F
on the normal.

108. The angle MNC=LCN=LCN, if I be the point


where the tangent meets the other asymptote,
.*. MN is parallel to Clj and passes through P the middle
point of LI.

109. The diameter of curvature being the same for


both, it follows that /SP : S'P is a constant ratio.
110.

CK'^=CF^ + PK'^ + 2PK\PF=AC'^-hBC^ + 2AC.BC;


110 Miscellaneous Problems,

111. If P, V, R, Q be the points of contact of AB,


BC, CD, DA,
2ASB = PSV-\-FSQ,
and two right angles
= P/S" V-h ASP + VSC= PSV+ ASQ + CSR ;

:.PSV=QSE = 2QSD,
and 2ASB = 2QSD + PSQ = 2ASD,
ASB = ASD = right angle.
.'. 2l

Also ASP + DSE = ASD,


.'.
PSQ is a straight line and PA, ED intersect on the
directrix.

112. If the tangent at P meet the director circle in E


and T, perpendiculars to the tangent through E and T are
tangents to the ellipse.
Draw PB parallel to CE, meeting CT in V and take
CQ^= CV. CT', similarly find the point D on CE ;

then andCQ CD
are conjugate diameters, and the con-
struction is completed in Art. 216.

113. Q'E' meets the axis in T, the pole of NP ;

.*. the tangents at Q', E', meet at a point on E NP,


LetCPmeet Q'PMn F;
then EN.PN=^EN^-EN.EP = EC''-CN'^-CY. CE,
= EC.CY- CN^ =CA'- CN' = PNK
/. EN PN=PN PN=AC BC=Q'M QM,
: : : :

and the tangent at Q passes through P'.


114. If S^ be the other focus of the fixed ellipse, aiKi
H of the moving ellipse,
S'P = HP and S'Q^-HQ.
Join S'H meeting the chord in Z, and let fall SY the
perpendicular on the chord ;
then SY.S'Z^^SY. HZ^BC\ and in the chord touches
a confocal conic.

115. be the centre of the circle,


Let a chord of PQ
intersection not perpendicular to the axis, meeting an
asymptote in L, and the axis in K,
Micsellaneous Prohlenis, 111

angle EOG^ 90« - LKG= 90" - LGA + CLK


= 46<^ + CLK^ LGA + LGE= EGO,
/. EGO is isosceles, and E lies on a fixed line perpendicular
to the axis.

116. Project the ellipse into a circle, and prove that


the angle DQP=QPR,
observing that the tangent and
common chord are equally inclined to the axis.

117. Reiprocates into the following :

If be a fixed point, and SK a tangent to a circle


/S'

centre G, and if TE be any other tangent from a point T,


and the angle GTE=GSK, the locus of T is a circle pass-
ing through S,

118. HZ being perpendiculars on the tangent,


SY,
Pq HZ PT TZ TR TG PR PE;
: :: : :: : :: :

/. Pq PR HZ AG HP,£G AG. GJJ


: :: : :: :

HP.PG GB' PG S'P; :: : :: :

.'.
Rq is parallel to SG.

119. IfPTfQ T and pt, qt be two near positions and tMy


Tm be drawn perpendicular to PT^ Qt,
then tM Tm in the ratio compounded of PT QT or
: :

GD GE and Pp.QO' Qq,PO or Pi^ QP' Q0\ PO


: : : :

being the radii of curvature.

Hence tM=Tm, or the normal at T to the locus of T


bisects PTQ.
Therefore T lies on a confocal ellipse.

120. Referring to the figure of Art. 148, and drawing


the lines, ADOG;
a circle can be drawn through

angle DOA
.-.
= DGA = 90' -GVA =AEG,
and the triangles AOD, AGE are similar.
AO AD GE AG, .-. : :: :

or AO,AC-=AD.A'D'-^BG\
112 Miscellaneous Frohlems.

121. Referring to the preceding theorem, describe a


sphere centre G and radius GA ; the tangent from any
point of the ellipse to this sphere will be equal to the
tangent from P
to the circle of curvature.

Describing a similar sphere with centre G\ the sum of


the tangents = FA' = SS\

122. In the second figure of Art. 144, take smj point T


in the tangent at P
and let C be the centre of the upper
sphere.
Then CRP and CSP are right angles, PR = PS, and
TR — TS these lines being tangents.
.'. T lies in the plane through CP perpendicular to the
plane CRPS-,
.-. angle SPT^RPT, and RTP = STP;
similarly RTP = S'TP.
Hence RTR' = STP + S'TP = STP + STQ^PTQ,
TQ being the other tangent.

123. Produce ER to E' making RE' equal to RE.


Then the polar of E' passes through E.
Now C is the pole of PQ which passes through E.
Hence CE' is the polar of E and is therefore parallel to
ARB.
Hence CE is bisected by AB.
Again CE bisects in E the polar of E' which is parallel
to^^.
Therefore CE bisects AB,
Therefore ACBE is a parallelogram.
124. Let be the centre of the conic which touches
the sides AB, BC, CD, DA in E, F, G, H.

Then since OA, OB, OG, OD bisect HE, EF, FG, GH


the sum of the areas of the triangles AOB, COD is half the
area of the quadrilateral
Miscellaneous Problems. 113

Let AB, CD meet in K and let O, 0' be two positions


of a
Draw OM, OM'
perpendicular to AB and ON, O'N' per-
pendicular to CD.
OK, O'L perpendicular to OM and ON:
Also draw
then OM.AB + ON, CD = 0'M\AB + 0'N\ CD,
Hence OK. AB = 0L. CD.
Therefore OL : OA^'in a constant ratio.
Hence the locus is a straight line.

125. By Art. 241 the sum or difference of the tangents


isproportional to the distance between the ordinates of the
points where the circles touch the curve according as the
point does or does not lie between those ordinates.

126. If SY, S^ Y^ be the perpendiculars from the focus


on the tangent at P, CY' is parallel to SP and, if is ;
DK
the perpendicular on CY',
DK : OD = SY SP = BC : :: CD;
DK=BC..'.

127. If Fand 7^ are contiguous corners of the parallelo-

gram formed by the tangents, and if CV and CT meet in


F
E and the sides of the parallelogram formed by the
points of contact,
CE. CV= CP' and CF. CT=CD';
.'.
(CE. CF) (CV. CT) = {CP CDJ. .

128. Taking the figure of Art. 12, let TL, TM, TN be


drawn perpendicular to SF, SP, FF\ and let the circle
intersect FF' in G and G'\

then TG : TN :: TL : TN :: SM TN :

:: SA :
AX,
.'. TG is parallel to an asymptote.

129. For 7W bisects QSq, Art. 12, and FS bisects the


outer angle, Art. 5.

B. 0. s. 8
114 Miscellaneous Problems.

130. Let the chord Qq, normal at g, meet the directrix


in F, and
let T
be the pole of Qq then S being the pole ;

of the directrix, ST
is the polar of F, and therefore T
being a point in the directrix, is a right angle. TSF
Taking V
as the middle point of Qq^ let meet in FS L
the polar of V, which is parallel to Qq.
Then LSQ = TSQ - TSL = TSq - TSF
=FSq=SqV-SFq
= Pq V-STq, V T, S, q, V are concyclic,
= PqV-QTV,KYi.m,
=:PVq-QTV=^QVN-QTV
^TQV=LTQ, V 7Z,e Fare parallel.
.•. L, T^ S, Q are concyclic, and
TQL=TSL = 90^.
131. (1) If the plane through the axis and the given
point P
intersects the cone in VA, VB, describe a circle
passing through P
and touching VA, VB; then if APB
is drawn touching the circle, AB
is the axis of the sectien
of which P
is a focus.

(2)Produce VP to Q making PQ = PF, and in the


plane above mentioned, draw parallel to VB, meeting VK
VA in ^, and VL parallel to VA, meeting VB in Z;
Then APL is the axis of a conic of which P is the
centre.

132. If aS',
S' are the foci,, and X
the foot of the
directrix, VS'S is a straight line, and XSS' is an isosceles
triangle.

Taking A and A^ as the corresponding vertices, draw


AL and A'L^ parallel to SS\ meeting AS' inL and AS
inZ'.
The latera recta are in the ratio of VS to VS',
and VS AS=A'L' ^Z' and VS' A'S'=AL A'L.
: : : :

Now AX=
LX, A'X= rX, and = AL'; AL
but VS. AL'== AS. AT
^nd VS\A'L = A'S'.AL;
.'. VS : VS' = AS.AX : A'S'AX.
CAMBRIDGE: PRINTED BT C. J. CLAY, M.A. AND SONS, AT THE UNIVERSITY PRESS.
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