过双曲线x^2/3-y^2/6=1的右焦点F2,倾斜角为30°的直线交双曲线于A,B两点,求绝对值AB
问题描述：
过双曲线x^2/3-y^2/6=1的右焦点F2,倾斜角为30°的直线交双曲线于A,B两点,求绝对值AB
答
a^2=3
b^2=6
c^2=a^2+b^2=3+6=9
c=3
右焦点坐标是(3,0)
k=tan30=√3/3
所以直线方程是
y-0=√3/3(x-3)
y=√3/3(x-3) 代入双曲线方程得
x^2/3-[√3/3(x-3)]^2/6=1
2x^2-[1/3*(x^2-6x+9)]=6
6x^2-x^2+6x-9=18
5x^2+6x-27=0
xa+xb=-6/5
xaxb=-27/5
(xa-xb)^2=(xa+xb)^2-4xaxb
=36/25+108/5
=576/25
(ya-yb)^2=(√3/3)^2(xa-3-xb+3)^2
=1/3 * 576/25
=192/25
|AB|=√[(xa-xb)^2+(ya-yb)^2]
=√(576/25+192/25)
=16√3 / 5