Plane and solid analytic geometry; an elementary textbook . Fig. 88. Ch. XII, § 86] POLES AND POLARS 161 Let the coordinates of the point P2 where CP1 cuts thehyperbola be (xv ya). Then the equation of the tangent 2 k b2x2x — a2y2y = a2b2, and the equation of the polar of Px isb2xxx — a2yxy = a2b2. But since Px and P2 are on the same line through the x xorigin, — = —-, and these lines are evidently parallel. Let the student prove the same theorem for the ellipse. 3. The polar of any point Px with respect to a parabola isparallel to the tangent at the point where a diameter throughP1 cuts the p
Plane and solid analytic geometry; an elementary textbook . Fig. 88. Ch. XII, § 86] POLES AND POLARS 161 Let the coordinates of the point P2 where CP1 cuts thehyperbola be (xv ya). Then the equation of the tangent 2 k b2x2x — a2y2y = a2b2, and the equation of the polar of Px isb2xxx — a2yxy = a2b2. But since Px and P2 are on the same line through the x xorigin, — = —-, and these lines are evidently parallel. Let the student prove the same theorem for the ellipse. 3. The polar of any point Px with respect to a parabola isparallel to the tangent at the point where a diameter throughP1 cuts the parabola. Y. Fig. 89. We may let the coordinates of P2 be (xvy-[)> Thenthe equation of the tangent at P2 is yxy = mx + mxv andthe equation of the polar of P1 is yxy = mx + mxv Theseequations are seen at once to represent parallel lines. 162 ANALYTIC GEOMETRY [Ch. XII, § 86 These two theorems show that the polar of a point on adiameter is one of the system of parallel chords bisectedby that diameter. 4. If the line joining the centre C of any central conic toany point Px cuts the conic in P2 and the polar of P1 in P3,then CP1 x CP3 = OP*. We shall give the proof for the hyperbola, usingFig. 88. The equation of QPX is y = — x. The coordinates of P2, where this line cuts the hyperbola are found to be ahxl ~~A <%1 and -\Jb2x^ — a2y-f ^/b2xx2 — a2y^ and the coordinates of P3, where it cuts the polar, b2xxx — a2yxy — a2b2, a2b2xx , a2b2y1 b2xx2 — a2yx2 b2xx2 — a2yx2 Hence CPX = ^x2 + y2, ^2~ V b2x2-a2y2 _g%^X* + y* 3 b2x2 - a2y2 From these values we see at once that CPX x CP3 = OP2.
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