问题 填空题
已知双曲线
x2
a2
-
y2
b2
=1(a>0,b>0)的离心率e=2,过双曲线上一点M作直线MA,MB交双曲线于A,B两点,且斜率分别为k1,k2.若直线AB过原点,则k1•k2的值为______.
答案

设M(x,y),A(x1,y1),B(-x1,-y1),则k1=

y-y1
x-x1
,k2=
y+y1
x+x1

∴k1•k2=

y-y1
x-x1
y+y1
x+x1
=
y2-y12
x2-x12

x2
a2
-
y2
b2
=1,
x12
a2
-
y12
b2
=1

∴两式相减可得

x2-x12
a2
-
y2-y12
b2
=0

y2-y12
x2-x12
=
b2
a2

∵双曲线的离心率e=2,

a2+b2
a2
=4

y2-y12
x2-x12
=
b2
a2
=3

∴k1•k2=3

故答案为3.

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