veniently be found by producing XE to y, making Ey= EY, and on Xy describing a semi-circle cutting Ed perpendicular from E to Xy in d. Ed is the required mean proportional.] Then a circle described with centre C and radius CD cutting CA in A and A, will determine A and A, the extremities of that diameter, and if Cd, be made = ND on CA1, and a parallel to d1n be drawn through A, cutting CB in B, this will determine an extremity of the other. The curve can then be completed by preceding problems.

1

Proof. The construction depends on the known proposition that EP. EP, EQ. EQ, CB2: CA; PP, and QQ, being any : :: chords parallel to the conjugate diameters CB, CA and intersecting in E. Admitting this, then by Prop. 4, p. 110,

QN2 : AN.NA, :: EP.EP1 : EQ . EQ,.

By the construction

and

but

...

AN. NA, QN° :: EQ. EQ, : EP. EP1,

which proves that AA, is the diameter parallel to EQQ, ·

Also by construction

CB: CA, :: QN : ND;

2

CB2: CA,2:: QN2: AN. NA1,

or CB is the semi-diameter conjugate to CA1.

That EP. EP, EQ. EQ, :: CB: CA' may be proved thus:-
Through E draw the diameter ERR, and draw the ordinate

RU parallel to PP, or to CB, then by Prop. 4, p. 110,

and

so that

but

or

...

CB2 – RU2 : CU2 :: CB2 – PM2 : CM2;

RU2: CU2 :: EM2: CM2;

CB2 : CU2 :: CB2 – PM2 + EM2 : CM2,

CB2 : CB2 – PM2 + EM2 :: CU2 : CM2 :: CR2 : CE2;

113 GIVEN CENTRE, DIRECTION OF MAJOR AXIS & 2 TANGENTS.

or

Similarly

or

:

CA2 : EQ. EQ, :: CR2 : ER. ER1,
EP. EP, : EQ. EQ, :: CB2 : CA2.

PROBLEM 68. To describe an ellipse, the centre (C), direction of major axis CT, and two tangents (PT, Pt) being given (Fig. 67). Bisect the angle TPt between the given tangents by PR meeting CT in R, and draw PU perpendicular to PR meeting CT in U.

Describe a circle round the triangle RPU and draw a tangent from C to this circle meeting it in K. CK will be the distance of either focus from C, i.e. make CF = CF1 = CK, and F and F will be the foci of the required ellipse. From F draw FY perpendicular to Pt meeting it in Y, and make YL on FY produced

YF. Draw FL cutting Pt in Q, and Q will be the point of contact of Pt, i.e. Q will be a point on the ellipse, which can therefore be completed by preceding problems.

Proof. Since CK is a tangent to the circle RPU,

.::

or

CK : CR :: CU : CK;

CK+CR : CK-CR :: CU + CK : CU - CK,
FR: RF, :: FU : F1U,

i.e. FU is divided harmonically in R and F, or P{FRF1U} is

1

a harmonic pencil. But the angle RPU is a right angle, and therefore PF and PF, make equal angles with PR (p. 15). Therefore also the angle FPQ = the angle FPT since PR bisects the angle QPT; or the tangents from P make equal angles with the focal distances of P: a known property of the ellipse.

FL is evidently the length of the major axis, for, by the construction QL = FQ, and therefore FL = F1Q+QF, the sum of the focal distances (Prob. 60, p. 101).

1

It follows that Y is on the auxiliary circle, for CF CF, and FY YL; therefore CY is parallel to and equal to FL- CA: and similarly if FY,, FZ and FZ, are perpendiculars from the foci on the tangents, Y, Z and Z, are all on the auxiliary circle. Produce YF to meet the auxiliary circle in Y2, then FY, is equal to F, Y,, and therefore

1

Similarly

i.e.

FY. F‚Y1 = FY.FY2 = AF. FA ̧. (Euc. III. 35.)
FZ. F ̧Z1 = AF.FA1=FY.F ̧Y,,

FY: FZ :: F,Z,
FZ, FY1;

: 1

and since the angle YFZ is equal to the angle Y1F1Z1, therefore the triangles YFZ, Z,F, Y, are similar (Euc. v1. 6), i.e. the angle FZY = F1Y,Z1.

1

Circles can be described about the figures

YFZP and FZ1PY1,

and therefore the angle FPY = the angle FZY,

therefore the angle FPY = the angle F1PZ1, which proves the property above referred to.

PROBLEM 69. To describe an ellipse, the centre C, the directions of a pair of conjugate diameters CT, Ct, a tangent Tt, and a point P being given (Fig. 68).

[P must lie between the line Tt and a parallel corresponding line on the other side of C.]

Draw PCL meeting Tt in L, and make CP, CP. P, is a point on the curve.

Take a mean proportional (Lm) between LP and LP,

and

120 GIVEN CENTRE, DIREC. OF CONJ. DIAM. TANGENT & POINT.

make LM on LT equal to Im. On Tt describe a semicircle Tqqt; draw Mn perpendicular to LM and make Mn=CP. Draw Ln cutting the semicircle in q and q,

From q

draw q

perpendicular to LT meeting it in ; then Q will be the point of contact of Tt, and Qq will be the length of CD the semi-diameter conjugate to CQ; the curve can therefore be completed by Problems 62 or 63.

Since In cuts the semicircle in two points, there are two solutions.

Proof. The construction depends on the property of the ellipse proved in Problem 67, that the rectangles contained by the segments of intersecting chords are in the ratio of the squares of the parallel diameters; and on the further property that if the tangent at Q meet a pair of conjugate diameters in T and ₺, and CD be conjugate to CQ,

QT. Qt CD.

=

If Q be the point of contact of Tt it follows that

LP. LP, : LQ2 :: CP2 : CD2;

but LP. LP, LM by construction,

CP:CD;

.. LM: LQ :: CP : CD;

but by construction

LM: LQ :: Mn : Qq,

and Mn = CP, so that Qq = CD; also Qq = QT. Qt, since Tqt is a semicircle, therefore CD2 = QT. Qt in the figure as drawn.

To prove that it does so in the ellipse, draw the ordinates QN, DK, parallel to Ct, and let the tangent at D meet CT in R, then by similar triangles,

PROBLEM 70. To describe an ellipse, the centre C, two tangents PT, QT, and a point on the curve (R) being given (Fig. 69).

[It is of course possible to draw at once two more tangents by producing TC to T,, making CT, = CT, and drawing through T parallels to TP, TQ. The point R must lie within the quadrilateral thus formed. Let the parallel (T,t) to TP meet TQ in t.]