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GIVEN TWO TANGENTS AND THREE POINTS.

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Find a mean proportional (Lk) between PL and PL,. (If PL, be made equal to PL, P, will be a point on the curve.) On PP, make PK= Lk, then QK will intersect TV in D, the extremity of the diameter TV.

On TV take a point C such that TC : CD :: CD : CV; and if, as in the figure, Q and R are on opposite branches of the hyperbola, C must be taken between T and V; i. e. on TV as diameter describe a semi-circle; draw DM making an angle of 45° with DV and meeting the semi-circle in M, and from M draw MC perpendicular to TV. Evidently MC is a mean proportional between CT and CV, and is equal to CD. C will be the centre of the hyperbola, and the asymptotes can easily be determined and the curve completed by preceding problems.

The proof is identical with that for the ellipse.

PROBLEM 112. To describe an hyperbola, two points A and B on the curve and three tangents PQ, QR, RP being given (Fig. 106). [Either no one of the three tangents must pass between the

points or all three must do so, and the points must not lie within the triangle formed by the tangents.]

Draw a line through AB cutting the tangents through P in L and M and the remaining tangent in N.

Find X the centre and D, D, the foci of the involution A, B and L, M (Prob. 13). D or D, will be a point on the chord of contact of the tangents PL, PM.

[In the figure La on PQ = LA, and Bm on a parallel to PQ=BM ; am cuts AB in X, the required centre, and XD = XD1 = a mean proportional between XM and XL.]

Similarly, find X, the centre and E, E, the foci of the involution A, B and M, N (Prob. 13), and E or E, will be a point on the chord of contact of the tangents RM and RN.

Find MV the harmonic mean between ME and MD, M being the point of intersection of AB with the given tangents which has appeared in each of the above involutions, then QV (Q being the intersection of tangents through N and L) will meet the tangent through M in its point of contact (g) with the curve.

Therefore qDr will be the chord of contact of the tangents PQ, PR, and Eqp the chord of contact of the tangents RP, RQ. The proof is identical with that for the ellipse, p. 135.

PROBLEM 113. To describe an hyperbola, two tangents TP, TQ and three points A, B, C on the curve being given (Fig. 107). [The points A, B, C being taken together in pairs, each pair

184

GIVEN TWO POINTS AND THREE TANGENTS.

of points must be either both on the same side or both on opposite sides of both tangents. In the figure A and B are both on the same side, and B and C on opposite sides of both TP and TQ, as also C and A.]

Draw the line AB cutting the given tangents in P and Q. Find X the centre and E, E, the foci of the involution A, B and P, Q (Problem 13).

[In the figure Pa PA and BQ, parallel to Pa=BQ. Q,a cuts AB in X, the required centre. XE=XE1 = a mean proportional between XA and XB.]

E or E, will be a point on the chord of contact of the given tangents.

Again, draw BC cutting the given tangents in p and q, and find X, the centre and G, G1 the foci of the involution B, C and P, q. G or G, will be a second point on the chord of contact of the given tangents, the points of contact of which R, R, are therefore determined, and the problem reduces to several preceding.

Since E and E, can be joined to either G or G, four chords of contact can in general be drawn, so that there are four solutions. The construction depends on Prop. 7, p. 143.

PROBLEM 114. To describe an hyperbola, five tangents AB, BC, CD, DE, EA being given (Fig. 108).

[The pentagon formed by the given tangents must contain a re-entering angle.]

Draw AC, BD intersecting in F; and through the remaining angular point E of the pentagon draw EF meeting BC in P. P will be the point of contact of the given tangent BC. Similarly, if BD and CE intersect in G, AG will intersect DC in Q, the point of contact of the given tangent CD; and if CE and DA intersect in H, BH will intersect ED in R, its point of contact.

The problem therefore reduces to Problem 111, or the points of contact S and 7' of the remaining tangents can easily be determined.

The construction depends (as in the corresponding problem for the ellipse, p. 136) on Brianchon's theorem.

PROBLEM 115. To describe an hyperbola, five points ABCDE being given (Fig. 109).

Draw AB, DE intersecting in F, and BC, EA intersecting in G ; then, if FG meet CD in H, H will be a point on the tangent at A, which can therefore be drawn.

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GIVEN FIVE POINTS.

If a line be drawn through G and the intersection of AB and DC, meeting ED in K, K will be a point on the tangent at B. Hence two tangents with their points of contact being known and also (at least) one other point on the curve, the problem may be completed by Prob. 111, or the tangents at C, D and E may also be found by a similar construction to the above.

[If CD and EA intersect in L, and through L a line LM be drawn passing also through the intersection of BC and DE and meeting BD in M, M will be a point on the tangent at C'; and if LM meet AB in N, N will be a point on the tangent at D.] The construction of the tangent at E is left as an exercise for the student.

The construction (as in the corresponding problem for the ellipse, p. 138) is an adaptation of Pascal's theorem.

PROBLEM 116. To describe an hyperbola, four tangents AB, BC, CD, DA and a point E on the curve being given (Fig. 110).

Join EC and ED, cutting AB in c and D, respectively. Find I the centre and F and F, the foci of the involution A, c and B, D,

Fig.110.

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(Prob. 13); For F, will be a point on the tangent at E, which can therefore be drawn and the problem completed by Prob. 114.