设A(x1,y1)为椭圆x^2+2y^2=2上一点,F1,F2为此椭圆的两个焦点

答：
椭圆过A点的切线方程为
x*x1+2y*y1=2,比较斜率知直线l就是切线.
由点到直线的距离公式知
d=│0*x1+0*2y1-2│/√[(x1)^2+4(y1)^2]
x1^2+2(y1)^2=2带入化简,得
d^2=2/[4-(x1)^2]
│AF1│=a+ex1,│AF2│=a-ex1,
故│AF1│*│AF2│*d^2=[a^2-e^2*(x1)^2]*2/[4-(x1)^2]
=[2-1/2(x1)^2]*2/[4-(x1)^2]
=1
(√│AF1│*│AF2│)*d=1