Question

设直线l:y=x+1与椭圆x^2/a^2+y^2/b^2=1(a>b>0)

Answer 1

（1）把直线方程代入椭圆方程,整理得（a^2+b^2)x^2+2a^2x+a^2-a^2b^2=0,

△/4=a^2b^2(a^2+b^2-1)>0,.∴a^2+b^2>1.

(2)F(-1,0),b^2=a^2-1,由AF=2FB得x1+a^2=2(x2+a^2),

a^2=9/8,b^2=1/8.

8x^2/9+8y^2=1.