File:A first course in projective geometry (1913) (14773266561).jpg

Identifier: firstcourseinpro00smarrich (find matches)
Title: A first course in projective geometry
Year: 1913 (1910s)
Authors: Smart, Edward Howard
Subjects: Geometry, Projective
Publisher: London : Macmillian and Co.
Contributing Library: University of California Libraries
Digitizing Sponsor: MSN

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Fig. 69. The diagonals of this rectangle (which of course intersect atthe centre) are the asymptotes. We shall speak of BC as the length of the (pseudo-) conjugateaxis of the hyperbola, though B is not a point on the curve. Def. The hyperbola which has the same asymptotes, butBCB as transverse and ACA as conjugate axis, is called theconjugate hyperbola. It is confined to the other pair ofangles between the asymptotes and, as is easily seen, touchesat D and D, the above-mentioned parallelogram. V. The area of the triangle cut off from the asymptotes byany tangent is constant. First Proof. Let a be half the angle between the asymptotes 134 PROJECTIVE GEOMETRY (Fig. 70) and LR, LR be perpendicular to the transverse axis.Then CLcosa = CR, CLcosa = CR, and since LP=PL,CR + CR = 2CN.Hence CL-i-CL = 2CN seca. Similarly CL-CL = 2PN coseca. :. CL. CL = CN- sec2 a - FN- cosec^ a.
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But FN AN . NA ac^ PN2 Pig. 70.and AN. NA = AC2-CN2, BC^ „ sec^ a= tan-^ a — s— CN - AC2 AC^ cosec- a .. CN^ sec^ a - FN^ cosec^ a = AC^ sec^ aand CL. CL = AC^ sec^ a = constant. But the angle LCL is constant. .*. the triangle LCL is of constant area. Second Proof. Let LPL, MQM be tangents at two nearpoints F, Q on the hyperbola, cutting the asymptotes at L, L,M, M (Fig 71), and one another at T. Then LF=FL and MQ = QM. CARNOTS THEOREM 135 Also, since the opposite angles at T are equal, ALTM _ TL.TM _(PL+PT)(QM - QT)tUTTM ~ A LTMALTM-ALTM (PL-PT)(QM + QT)2(PT.QM-QT.PL) ALTM (PL-PT)(QM+QT) Now when P moves up to Q, PT and QT are ultimately vanishing quantities, PL and QM remaining finite. Hence the difference of the triangles LTM, LTM ultimately vanishes compared with either of them. That is to say, they are equal.

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