3 5 a 2 + 4 b 2 =1 a 2 = b 2 + c 2 ,所以a=3, b= 5 ,所以椭圆Γ的方程为 x 2 9 + y 2 5 =1 ; (Ⅱ)∵K=1,F(-2,0),∴设直线方程为y=x+2,A(x 1 ,y 1 ),B(x 2 ,y 2 ) 联立方程组
y=x+2
x 2 9 + y 2 5 =1 ,整理得14x 2 +36x-9=0, x 1 + x 2 =- 18 7 , x 1 x 2 =- 9 14 , ∴ |AB|= 2 | x 1 - x 2 |= 2 • ( x 1 + x 2 ) 2 -4 x 1 x 2 = 30 7 , 设O点到直线AB的距离为d,则 d= |0-0+2|