Take any second chord qq, parallel to QQ,, meeting FY in Y1 and the directrix in R1, let O, be its centre point, then since OYO,Y, it follows that the line 00, must pass through the

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OR OR

Ι

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point T, in which F, Y meets the directrix and is therefore fixed for all chords parallel to QQ1. This line T,O will pass through the centre (i.e. will be a diameter), because the chord through the centre parallel to QQ, is bisected by the centre and also by T2O. Let T1O meet the ellipse in P, and suppose qq, to move parallel to itself till it approaches and ultimately passes through P. Since 0,9 = 0,9, throughout the motion the points 9, 9, will evidently approach P, simultaneously, and in the limiting position 77, will be the tangent at P, It follows that if P. be the other extremity of the diameter through P,, the tangent at P, is parallel to QQ1, and therefore to the tangent at P..

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COROLLARY. The perpendicular on the tangent at any point from the focus meets the corresponding diameter in the directrix.

PROP. 3. If PCP, be a diameter and QVQ, a chord parallel to the tangent at P and meeting PP, in V, and if the tangent at Q meet PP, produced in T, then CV. CT=CP (fig. 60).

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Let TQ meet the tangents at P and P1 in R and r, and F being a focus draw RN perpendicular to the focal distance FP

meeting FP in N, rn perperpendicular to FP, meeting it in n, and RM, rm perpendicular to the focal distance FQ.

108

CONJUGATE DIAMETERS.

Let F, be the other focus and join F,P, F,P,.

Since CF CF, CP CP1, and the angle FCP the angle FCP, therefore the triangles FGP, FCP, are equal in all respects, and therefore the angle CPF= the angle CPF;

similarly

CPF, the angle CP,F,

and therefore the whole angle FPF1 = the whole angle F,PF. But the tangents are equally inclined to the focal distances, and therefore also the angle FPR = the angle F1P1r,

.. the angle FPR = the angle FP ̧r,

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i.e. the right-angled triangles RPN, rP ̧n are similar, and thereRP : rP、 :: RN : rn.

fore

But

But

or

CT-CP: CT + CP :: CP − CV : CP + PV,

.. CT : CP :: CP : CV,

. CT. CV = CP2.

Cor. 1. Since CV and CP are the same for the point Q1, the tangent at Q, passes through T or the tangents at the extremities of any chord intersect on the diameter which bisects that chord.

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Cor. 2. Since TP, : TP :: P‚V : VP, it follows that TPVp is harmonically divided (p. 13).

Let

The above proposition has been proved generally; it therefore holds when the diameter CP coincides with the major axis. P1 be any point on an ellipse (fig. 56) and draw the ordinate P1N perpendicular to CA, producing it to meet the auxiliary circle in p, and draw the tangent at P, meeting CA in T, then

and

CN.CT = CA2 = Cp2,

.. CpT is a right angle,

and therefore pT is a tangent at p to the auxiliary circle: hence

Cor. 3. The tangents at the extremities of corresponding ordinates of the ellipse and auxiliary circle intersect on the major axis.

Draw CD (fig. 56) the diameter conjugate to CP1, dDn the corresponding ordinate meeting the auxiliary circle in d, and the tangents at D and d meeting the major axis in t.

and

Then

P1N: ÎN :: BC: AC :: Dn: dn,

P1N: NT:: Dn : C'n,

since CD is parallel to PT,

pN: NT:: dn: Cn,

and therefore C'd is parallel to pT, i.e. pCd is a right angle, or

Cor. 4. Conjugate diameters in the ellipse project into diameters at right angles to each other in the auxiliary circle.

If the tangent at d meet the major axis in t, since dt is parallel to Cp, Dt (the tangent at D) will be parallel to CP„, or,

Cor. 5. If CD be conjugate to CP, CP, is also conjugate to CD. Since pCd is a right angle, the angle dCn is the complement of the angle pCN, and therefore equals the angle CpN, therefore the triangles CpN, dCn are equal in all respects, i. e. Cn=pN and dn = CN,

But

and

or

CP ̧2 = P1N2 + C'N2 and CD2 = Cn2 + Dn2,

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.. CP, 2+CD \2 = CN\2 ÷ pN \2 ÷ P1N ¦2 + Dn¦2

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AXES FROM CONJUGATE DIAMETERS.

PROP. 4. If PCP, DCD ̧ be conjugate diameters and QV be

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drawn parallel to CD meeting the ellipse in Q and CP in V, then

[QV is called an ordinate of the diameter PCP ̧.]

Let the tangent at Q (fig. 60) meet CP, CD in T and t, and draw QU parallel to CT meeting CD in U.

Then

But

Since

and

CV.CT=CP and CU.Ct= CD (Prop. 3).

CU = QV,

.. CD2 : CP2 :: QV. Ct : CV.CT

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PROBLEM 62.

the axes (fig. 61).

:: QV2 : CV. VT.

Ct: QV :: CT: VT,

CV. VT CV. CT – CV2

ཟ-

= C P2 – C V2 – PV . VP1,

QV2 : PV. VP, :: CD2 : CP2.

Given a pair of conjugate diameters to determine

PCP1, DCD, are the given conjugate diameters. Through D draw Q,DQ perpendicular to CP. Make DQ and DQ, each equal to CP and draw the lines CQ, CQ,. Then the major axis ACA, bisects the angle QCQ, and the minor axis (BCB) is of course a line through C perpendicular to ACA,. The axes are therefore determined in direction. To determine them in magnitude :On Q1C on opposite sides of C make Cq and Cq, each equal to CQ, then Qg will be the length of the major axis AA, and Q, Bisect each of these lines and

be the length of the minor BB. CA, CB will be given respectively.

191 will

Proof. Since CQ-Cq and BC bisects the angle QCq, therefore a, the point in which Qq cuts BC, bisects Qq and therefore Da is parallel

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to Q19 and
CA bisects Qq, and Db is parallel to Q,

= CA. Similarly b the point in which Qq, meets

and

1 =CB, and

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D, b, a are in the same straight line. Hence D is a point on the ellipse described with CA, CB as semi-axes. Also DQ is the normal at D, since Q is the instantaneous centre of rotation for the line ab moving along the axes. Therefore the tangent at D will be parallel to CP. Lastly, to shew that P will also be on the curve,

But

and

CQ2 + CQ ̧2 = 2CD2 + 2DQ2 (Euc. II. 12 and 13),

a known property of conjugate diameters. (See Cor. 6, p. 109.)

PROBLEM 63. To describe an ellipse having given conjugate diameters PCP,, DCD ̧.

This might of course be done by the last problem: the curve may however be drawn independently, though none of the following constructions give any information as to the position of the axes, foci, or directrices.