文[1]给出一个猜想,摘抄如下:猜想若A,B分别是椭圆x2/a2+y2/b2=1(a>b>0)一直径两端点,P为椭圆上任意一点(不与A,B重合),直线PA,PB与AB的共轭直径所在的直线分别交于C,D,则椭圆在点P处的切线平分线段CD.文[2]给出了上述猜想的证明,并把这一性质进行了再推广.本文中,笔者利用椭圆的参数方程来证明该猜 The text [1] gives a conjecture, which is extracted as follows: Suppose that if A and B are two ends of a diameter x2 / a2 + y2 / b2 = 1 (a> b> 0), P is any point on the ellipse A, B coincide), the straight lines PA, PB and AB conjugate diameter where the intersection of the intersection of C and D, then the ellipse at the point P of the tangent bisecting line segment CD [2] gives the proof of the above conjecture, And this nature has been re-extended.In this paper, I use the ellipse parameter equation to prove that the guess