Question

直线y=-x-1与椭圆x2/4b2+y2/b2=1交于A B两点 若OA垂直OB 求椭圆方程

Answer 1

A(x1,y1),B(x2,y2),OA垂直OB ,x1*x2+y1*y2=0,

y=-x-1,y1=-x1-1,y2=-x2-1,y1*y2=x1*x2+(x1+x2)+1,2x1*x2+(x1+x2)+1=0

椭圆x2/4b2+y2/b2=1,x^2+4y^2-4b^2=0,代y=-x-1入椭圆方程,

得5x^2+8x+(4-4b^2)=0,x1*x2=(4-4b^2)/5,x1+x2=-8/5,代入上式

2(4-4b^2)/5-8/5+1=0,b^2=5/8,