172 GIVEN CENTRE, TWO POINTS AND TANGENT.

PROBLEM 101. To describe an hyperbola, the centre C, two points (A and B) of the curve and a tangent Tt being given (Fig. 97).

[A second tangent can at once be drawn parallel to Tt on the other side of C and at the same distance from it; the points A and B must not lie between these lines.] If the given points lie on opposite branches of the curve, as e.g. A and B1, i. e. if they are on opposite sides of Tt, draw B,CB and make CB B will be on the same branch as A.

CB1, then

Then CT,

tangent in T, and Ct parallel to AB meeting it in t. Ct are the directions of a pair of conjugate diameters, and the problem reduces to Prob. 99.

In the figure CA1 = CA and ACA, meets Tt in L; Lm is a mean proportional between LA and LÁ ̧; LK: Lm :: OT : CA, where LK is perpendicular to Tt and OT = 1⁄2Tt.

Then a tangent (Kq) from K to the circle on Tt as diameter cuts Tt in Q, its point of contact with the curve. CD parallel to Tt and equal to Qq is the semi-diameter conjugate to CQ.

PROBLEM 102. To describe an hyperbola, the centre C, and three tangents (SV, VW, WS) being given (Fig. 98).

Through C draw TCT, meeting SV in T and SW in T,, so that TC = CT, (Prob. 14, p. 19). CT will be conjugate to CS. Draw

1

W

Then ES will cut Also PQ parallel to

Tv parallel to CV meeting VW in v, Tw parallel to CW meeting VW in w, and draw vT, wT, meeting in E. VW in P, its point of contact with the curve. vT will cut VS in Q, its point of contact, and QR parallel to CT or PR parallel to Tw will cut WS in R, its point of contact. The problem can be completed by several of those previously given or thus ;

Draw QN parallel to CT meeting CS in N. QN is an ordinate of the diameter CS, and therefore CA, the length of the semidiameter, is a mean proportional between CS and CN (Prop. 4, p. 164). Similarly, if Qn be drawn parallel to C'S meeting CT in n, CB must be taken as a mean proportional between Cn and CT. PROBLEM 103. To describe an hyperbola, the centre C and three points P, Q, R being given (Fig. 99).

[Each of the points must lie either between both pairs of lines furnished by the remaining points and their corresponding points, or outside both these pairs of lines.]

174

GIVEN CENTRE AND THREE POINTS.

Bisect PQ in p, QR in q, and RP in r, and draw Cp, Cq and Cr, producing each indefinitely. PQ is a double ordinate of the

Fig.99.

R

T

diameter Cp, and therefore the tangents at P and Q will intersect on Cp; similarly the tangents at Q and R will intersect on Cq, and at R and P on Cr. If therefore a triangle be drawn (Prob. 15, p. 20), the sides of which pass through P, Q and R, and the vertices of which lie on Cp, Cq and Cr respectively, the sides of this triangle will be the tangents at P, Q and R. Take any

point a on Cr; draw Pa, Ra cutting Cp, Cq in b and c respectively; join be cutting PR in x, and draw a cutting Cb in T. QT, PT will be the tangents at Q and P respectively; and if PT meet Cr in t, Rt will be the tangent at R. The problem may be completed by preceding problems.

PROBLEM 104. To describe an hyperbola, the foci F, F, and a point P on the curve being given (Fig. 100).

It has been shewn already that the difference between the focal distances of any point on the curve is equal to the transverse axis (p. 153).

1

Let FP be greater than FP. On PF, make Pf=PF. Draw FF,, bisect it in C and make CA=CA, F, AA, will there= {F1ƒ.

fore be the vertices of the curve, and the problem reduces to Prob. 89.

PROBLEM 105. To describe an hyperbola, the foci F, F, and a tangent Tt being given (Fig. 100).

[The tangent must lie between F and F1.]

From For F, drop

Bisect FF, in C, the centre of the curve. a perpendicular (as FY) on the tangent. On CF, CF, make CA=CA1=CY. A and A, will be the vertices of the curve, and the problem reduces to Prob. 89.

1

PROBLEM 106. To describe an hyperbola, a focus F, a tangent PT with its point of contact P, and a second point Q on the curve, being given (Fig. 100).

If F and Q are on the same side of PT, the solution has already been given in the corresponding problem for the ellipse (Prob. 76). Hence the case of F and Q lying on opposite sides of PT need alone be considered here.

From F draw FY perpendicular to PT meeting it in Y, on FY produced make Yƒ= YF; then Pf will be a locus of the second focus. From ƒ on ƒP, on either side of ƒ, make ƒq = FQ. Bisect Qq in r, and draw rF, perpendicular to Qq meeting Pf in F1, which will be the second focus. Hence, both foci being known, the problem may be completed by Probs. 104 or 105.

Since q may be taken on either side of ƒ, there are in general two solutions.

176

GIVEN FOCUS, TANGENT AND TWO POINTS.

Proof. That fP is a locus of the second focus has been shewn in p. 157; that the second focus is at the intersection of ƒP and rF, is evident thus :-it must be so situated that

If F and Q are on opposite sides of PT, two hyperbolas can in general be drawn.

If F and Q are on the same side of PT, and the distance of Q from F is greater than its distance from the line drawn through ƒ perpendicular to Pf, two hyperbolas can in general be drawn.

If F and Q are on the same side of PT, but the distance of Q from F is less than its distance from the above perpendicular, one hyperbola only can in general be drawn, but an ellipse can also be drawn.

If F and Q are on the same side of PT, and the distance of Q from F is equal to its distance from the above perpendicular, a parabola can be drawn fulfilling the required conditions, but no hyperbola or ellipse, since the second focus removes to an infinite distance.

PROBLEM 107. To describe an hyperbola, a focus F, a tangent RT and two points P and Q of the curve being given (Fig. 101). If F, P and Q are all on the same side of RT, the solution has already been given in Prob. 77, the corresponding problem for the ellipse. Hence the cases of one or both of the points P, Q lying on the opposite side of RT to F need be considered.

Case 1. Let F and P be on the same side of RT, and Q on the opposite side. Produce PF to q, make Fq= FQ, and with centre P and radius Pq describe a circle qG.

From F draw FY

perpendicular to RT, produce it to ƒ, and make Yƒ= YF. With centre ƒ and radius FQ describe a circle GH, and find the centre (F) of a circle touching the circles qG and GH internally and