题目内容：

点P是椭圆x^2/a^2+y^2/b^2=1(a>b>0)上的动点,点A,B关于原点对称.求证:kPA·kPB为定值

优质解答

设直线AB的方程为Ax+By=0,点A(x1,y1)点B(-x1,-y1),P(x0,y0)则kPA*kPB=(y0-y1)(y0+y1)/[(x0-x1)(x0-x1)]=(y0^2-y1^2)/(x0^2-x1^2)又因为x^2/a^2+y^2/b^2=1(a>b>0),所以(x1,y1),(x0,y0)都满足y^2=b^2*(1-x^2/a^2)解得k...