and combining (6) and (7), we get

RN2 – EN2 – BC2 = RE. Er = RE, Re,

where r is the point on the other branch of the conjugate hyperbola corresponding to R, and e is the point in which Rr meets the other asymptote.

To draw a tangent and normal at any point of the curve (Fig. 88).

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Let P1 be a point on the curve adjacent to any point P, and let the chord PP, meet the directrices in K and K1. Draw KF to the corresponding focus: then FP: FP, :: PK : P,K, or PK bisects the exterior angle between PF and PF produced (Euc. VI. prop. A). Hence, exactly as in the case of the ellipse (p. 102), when P, moves up to and coincides with P, so that the chord PP1 becomes the tangent at P, the line FK becomes perpendicular to the line FP drawn from the focus to the point of contact of the tangent. The tangent at any point P of an hyperbola may therefore be drawn by drawing a line from P to either focus, erecting a perpendicular to this line at the focus meeting the directrix, and drawing the tangent through this point and the proposed point of contact. It may also be drawn by making use of the known property that it bisects the angle between the focal distances. For in the two triangles PFK, PFK,

FP : PK :: FP : PK1,

and the angle PFK = the angle PF,K,, each being a right angle, .. the angle FPK=F,PK1. (Euc. vI. 7.)

Hence the normal bisects the exterior angle between the focal distances.

PROBLEM 90. To describe an hyperbola, an asymptote CD, a focus F, and a point P being given (Fig. 88).

From F draw FD perpendicular to CD, then D will be a point on the directrix as has been previously proved. Through P draw Pf parallel to CD and make Pf= PF, then ƒ will be a second point in the directrix, which is therefore determined. Draw CF

158

GIVEN ASYMPTOTE, FOCUS AND POINT.

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perpendicular to Df meeting the given asymptote in C, which will evidently be the centre of the curve. Make CA, CA, on CF each equal CD and AA, will be the transverse axis, and the curve is completely determined.

Since Pf can be measured on either side of P there are generally two solutions.

Proof. The only step in the construction which is not obvious is taking f as a point on the directrix. It can easily be shewn to hold in the hyperbola, for draw Am parallel to the asymptote meeting the directrix in m, then in the hyperbola

FP FA:: PM : AX

:: Pƒ : Am,

where PM is a perpendicular on the directrix.

But

Am = DL = AF, .. FP=Pƒ;

and conversely, if Pf be made = PF, ƒ will be a point on the directrix.

PROBLEM 91. To describe an hyperbola, an asymptote CT, a tangent Tt, and a focus F being given (Fig. 89).

From F draw FD perpendicular to CT meeting it in D, and FY perpendicular to Tt meeting in Y. Then D and Y are points on the auxiliary circle. Bisect DY in K and draw KC perpendicular to DY meeting CD in C. C will be the centre of the curve, CF the direction of the transverse axis, and CD or CY its semi-length.

PROBLEM 92. To describe an hyperbola, an asymptote CD, a directrix DD, and a point P being given (Fig. 89).

From D draw DF perpendicular to CD. DF will be a locus of the focus. Through P draw Pf parallel to CD meeting DD1 in ƒ, and with centre P and radius Pf describe an arc cutting DF in F. F will be a focus, and FC drawn perpendicular to DD, will intersect CD in C, the centre of the curve; which is therefore completely determined.

[The problem is exactly the converse of Prob. 99.]

PROBLEM 93. To describe an hyperbola, the asymptotes CD, CD1 and a point P on the curve being given (Fig. 89).

Bisect the angles between the asymptotes by the lines ACA,, BCB1; then ACA, in the angle in which P lies is the position of

perpendicular to AA, Take a mean propor

the transverse axis. Through P draw QPq and meeting the asymptotes in Q and q. tional, as Pb, between PQ and Pq. Pb will be the length CB of the conjugate semi-axis.

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Draw BE parallel to 44, meeting the asymptote in E; then BE is the length CA of the transverse semi-axis and CE = CF, the distance of either focus from C.

Proof. The only step in the construction requiring demonstration is that in the hyperbola BC2 = PQ. Pq.

Let N be the foot of the double ordinate Pp; by similar triangles CAE, CNQ.

2

QN2: AE2 :: CN: AC and AE= BC,

.. QN2 – BC2: BC2 :: CN2 – AC2: AC2;

but CN2-AC2=AN. NA, and (p. 156) PN2: AN. NA, :: BC2: AC,

160

GIVEN ASYMPTOTES AND TANGENT.

··. QN2 – BC2 = PN2 or QN2 – PN2 = BC2,

i.e. (QN+PN) (QN − PN) = BC2 = PQ . Pq,

for by the symmetry of the curve Nq= NQ.

If

curve.

qp be made equal to QP, p will evidently be a point on the

PROBLEM 94. To describe an hyperbola, the asymptotes CT, Ct, and a tangent Tt to the curve being given (Fig. 89).

Bisect Tt in P. P will be the point of contact of Tt, i. e. will be a point on the curve, and the problem therefore reduces to Problem 93.

The proof will be found on p. 163.

DEFINITION. Any straight line drawn through the centre and terminated both ways either by the original curve or by the conjugate hyperbola is called a diameter, and by the symmetry of the curve every diameter is bisected by the centre. A diameter CD parallel to the tangent at the extremity of a diameter CP is said to be conjugate to CP.

The following important properties of the hyperbola should be carefully noticed.

PROP. 1. If from any point Q in an asymptote QPpq be drawn meeting the curve in P, p and the other asymptote in I, and if CD be the semi-diameter parallel to Qq,

QP. Pq = CD2 and QP = pq (Fig. 90).

Through P and D draw RPr, DTt perpendicular to the transverse axis, and meeting the asymptotes in R, r and T, t; let Rr meet the axis in N.

Then

and

But

QP RP :: CD : DT

:

Pq: Pr CD: Dt by similar triangles,

::

)

Similarly

qp ⋅ pQ = CD2 = QP .Pq;

or, if Vbe the middle point of Qq,

QV2 – PV2 = QV2 − pV2.

Hence PV=pV, and therefore PQ=pq.

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COR. If a straight line PP,pip meet the hyperbola in P, p, and the conjugate hyperbola in P1, P1, PP1 =PP1

1

For if the line meet the asymptote in Q, 7,

QP1=P1q and PQ = qp, .. PP1=PP1•

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PROP. 2. A diameter bisects all chords parallel to the tangents at its extremities, i. e. all chords parallel to its conjugate.

This can be proved exactly as in the analogous proposition for the ellipse.

Let QQ, (fig. 91) be any chord of an hyperbola meeting the directrix in R, and let O be the centre point of QQ, and ♬ the focus.

Join FQ, FQ1, and draw FY perpendicular to QQ1•