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本文介绍曲线Ax2+By2=C(AB≠0)的一条有趣性质,并以高考题为例说明其应用.1曲线的性质定理设曲线Ax2+By2=C(AB≠0)与直线P1P2相交于P1(x1,y1)、P2(x2,y2)两点,P为线段P1P2的中点,若直线P1P2、OP的斜率分别为k、m,则A+kmB=0.证明设P(x0,y0),则x1+x2=2x0,y1+y2=2y0,且xy00=1m.因
This article introduces an interesting property of the curve Ax2+By2=C(AB≠0), and uses the college entrance examination question as an example to illustrate its application. 1 The nature of the curve Theorem Let the curve Ax2+By2=C(AB≠0) cross the straight line P1P2. P1 (x1, y1) and P2 (x2, y2) are two points, and P is the midpoint of the line segment P1P2. If the slopes of the straight line P1P2 and OP are k and m, then A+kmB=0. Prove that P (x0, Y0), then x1+x2=2x0, y1+y2=2y0, and xy00=1m.