280 CONIC SECTIONS PROP. IV. The foot of the perpendicular from the focus on the tangent always lies on the circle described on the transverse axis as diameter. This proposition is proved in exactly the same way as Prop. iv. on the ellipse. PROP. V. The product of the perpendiculars from the foci on the tangent is constant. This proposition is proved exactly in the same way as Prop. v. on the ellipse, the only difference in the figure being that S, S lie without the circle instead of within as in the case of the ellipse. PROP. VI The tangent at any point of an hyperbola makes equal angles with the focal distances of the point. Let the tangent PZZ at the point P (fig. 33) meet the two directrices in Z, Z . Join ZS, SP, PS , S Z , and draw PMM parallel to the axis to meet the directrices in M, M . Then because SP : PZ = ePM : PZ = ePM : PZ = PS : PZ , and the angles PSZ, PS Z are right angles (Prop, iii.), therefore the triangles PSZ, PS Z are similar (Eucl. vi. 7). Therefore the angle SPZ = angle S PZ . Fig. 33. PROP. VII. To draw a tangent to an hyperbola at a point on the curve. First Method. Join SP, S P, and draw a line bisecting the angle SPS : this line is the tangent at P. (Prop, vi.) Second Method. Draw SZ at right angles to SP meeting the cor responding directrix in Z : ZP is the tangent at P. (Prop, iii.) Third Method. On SP as diameter describe a circle which will touch the circle on AA as diameter in some point Y : YP is the tangent at P. (Prop, iv.) PROP. VIII. To draw a pair of tangents to an hyperbola from an external point, First Method. Let (fig. 34) be the point, and S, S the foci. Fig. 34. Join OS ; with centre 0, and radius OS, describe a circle ; and with centre S and radius equal to AA describe another circle. It can be shown that these circles will always intersect in two points M, M . Join S M, S M cutting the curve in P, P . Then OP, OP will be tangents to the curve. Join SP, SP . Now S P-SP~AA -> S M = S P - MP, therefore SP = MP; and OS = OM ; therefore the two tri- Fig. 35. angles OPS, 0PM are equal in all respects, and the angle OPS = Therefore OP is a tangent to the hyperbola at P (Prop, vi.) Second Method Let O (tig. 35) be the point, and S, S the foci. Join OS, and upon it as diameter describe a circle cutting the circle described on AA as diameter (which it will always do) in Y, Y . Join OY, OY , and produce them if necessary to meet the curve in P, P . They will be tangents to the curve at P, P . Because OYS is a semicircle, the angle OYS is a right angle, and therefore OY is a tangent to the hyperbola (Prop, iv.) PROP. IX. If OP, OP be tangents to the same branch of an hyperbola at P, P and S be a focus, the angles OSP, OSP are equal : if OP, OP be tangents to different branches, the angles OSP, OSP are supple mentary. This proposition is proved in the same manner as Prop. ix. on the ellipse. PROP. X. If OP, OP be two tangents to the hyperbola, and S, S be the foci, the angles, SOP, S OP are equal. This proposition is proved exactly in the same way as Prop. x. on the ellipse. PROP. XI. If C be the middle point of A A , then CA 2 = CS . CX. .-. SA -SA:SA = A X-AX:AX. .-. AA :XX -SA:AX:1. A^ain SA + SA : SA = A X + AX: AX. SS :SA = AA :AX. .-. BS :AA -SA:AX-:1. From (1) and (2) a.) (2.) Also AA :XX =SS :AA CA:CX = CS:CA .-. CA 2 = CS.CX. CS:CX = CS 2 :OA 2 -CA 2 :CX 2 . Fig. 36. PROP. XII. If PN be an ordinate of the hyperbola, then PN 2 always bears a con stant ratio to AN .NA . It is proved exactly as in the case of the ellipse Prop, xii., that PN 2 :AN . NA =CB 2 : CA 2 . The ordinates of two hyperbolas which have the same transverse axis are in a constant ratio. Let PP N be the common ordinate of two hyperbolas, whose transverse axis is AA , and whose conjugate axes are CB, CB . Then PN 2 : AN . NA = CB 2 : CA 2 and P N 2 : AN . NA - CB 2 : CA 2 . Therefore PN S : CB 2 = AN . N A : C A 2 = P N 2 : CB 2 . .-. PN:CB = FN:CB or PN:P N = CB:CB . DEFINITION. The diagonals of the rectangle formed by the tangents to a hyper bola and its conjugate at their vertices are called asymptotes. PROP. XIV. If a straight line be drawn through nny point n one f tho asymptotes perpendicular to the transverse axis, meeting tho hyperbola in P, P and the other asymptote in Q (fig. 36), then QP. QP = B C S . From similar triangles QNO, EAO, we have
QN 2 :CN 2 = BC 2 : AC 2 ,