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定理1椭圆的中心为O,长半轴为a,短半轴为b,直线l交椭圆于P、Q,O到l的距离为d,则∠POQ=90°的充要条件是1/d~2=1/a~2+1/b~2.证明:设椭圆的方程:x~2/a~2+y~2/b~2=1(a>b>0),直线l的方程:y=kx+p,P(x_p,y_P),Q(x_Q,y_Q),解方程组:
Theorem 1 The center of the ellipse is O, the long half axis is a, the short half axis is b, the straight line l is elliptical to P, Q, and the distance from O to l is d, then the necessary and sufficient condition of ∠ POQ=90° is 1/ d~2=1/a~2+1/b~2. Proof: Set the equation of the ellipse: x~2/a~2+y~2/b~2=1(a>b>0), the straight line l The equations: y = kx + p, P (x_p, y_P), Q (x_Q, y_Q), the solution equations: