Question

已知椭圆C：x^2/a^2+y^2/b^2=1（a>b>0）的离心率为√3/2，

过右焦点F且斜率为k（k>0)的直线与C相交于A.B两点若AF=3FB,则k=？

Answer 1

e = √3/2 → c/a = √3/2 → b/a = 1/2
椭圆C: x^2/4b^2 + y^2/b^2 = 1 设直线 AB: my = x - c = x - √3b

设A（x1,y1),B(x2,y2)
AF/BF=3 → ly1/y2l = 3 直线AB代入椭圆 x^2+4y^2 = 4b^2

4b^2=(my+√3b)^2+4y^2=(m^2+4)y^2 + 3b^2 + 2√3bmy → (m^2+4)y^2 + 2√3bmy -b^2=0
可知 y1*y2<0 因此 y1/y2 = -3
那么 y1*y2 = -3y2^2 = -b^2/(m^2+4) → y2^2 = b^2/3(m^2+4)......1#
y1+y2 = -2y2 = -2√3bm/ (m^2+4) → y2^2 =3b^2m^2/(m^2+4)^2.....2#

1#=2# → b^2/3(m^2+4) = 3b^2m^2/(m^2+4)^2 →(m^2+4) = 9m^2 → m^2 = 1/2 → m =±1/√2

那么AB : y = ±√2(x-√3b), k =±√2