性质　椭圆的中心到对中心张直角的弦的距离为一定值 .具体对椭圆 x2a2 + y2b2 =1来说 ,直线 PQ交椭圆于 P,Q两点 ,且 OP⊥ OQ,O到 PQ的距离为 d,则 d2 =a2 b2a2 + b2 .证明　设 OP:y=kx,将它代入椭圆方程 x2a2 + y2b2 =1 ,得( b2 + a2 k2 ) x2 =a2 b2 ,∴ x2 =a2 b2b2 + a2 k2 The distance from the center of the property ellipse to the string at right angles to the center is a certain value. Specifically for the ellipse x2a2 + y2b2 =1, the straight line PQ intersects two points P and Q, and the distance between OP⊥ OQ and O to PQ is d, then d2 = a2 b2a2 + b2 . Prove that OP: y = kx, substituting it into the elliptic equation x2a2 + y2b2 =1, get (b2 + a2 k2) x2 = a2 b2, ∴ x2 = a2 b2b2 + a2 k2