Question

双曲线C：x2/a2-y2/b2=1的右焦点为F，过F且斜率为根号3的直线交C于A、B两点，若AF=4FB，则C的离心率为?

Answer 1

设AF=4m,BF=m.过A,B分别做准线的垂线，垂足为A1,B1。有双曲线定义得，
AA1=4m／e.BB1=m／e.过B做BD垂直于AA1垂足D。三角形ABD中，角ABD=30°，AD=0.5AB.即3m／e＝0.5×5m.解得e=6／5。

Answer 2

设A(c+4tcos60°，4tsin60°),B(c-tcos60°，-tsin60°),
A,B都在双曲线上，
∴(c+4tcos60°)^2/a^2-(4tsin60°)^2/b^2=1,①
(c-tcos60°)^2/a^2-(-tsin60°)^2/b^2=1,②，
②*16-①，16(c-tcos60°)^2-(c+4tcos60°)^2=15a^2,
16c^2-16ct+4t^2-(c^2+4ct+4t^2)=15a^2,
15c^2-20ct=15a^2,t=3b^2/(4c),
代入②，[c-3b^2/(8c)]^2/a^2-[3√3b^2/(8c)]^2/b^2=1,
(5c^2+3a^2)^2/a^2-27b^2=64c^2,e=c/a,
(5e^2+3)^2-27(e^2-1)=64,
25e^4+3e^2-28=0,e>1,本题无解.