题目内容
如图,过双曲线上左支一点A作两条相互垂直的直线分别过两焦点,其中一条与双曲线交于点B,若三角形ABF2是等腰直角三角形,则双曲线的离心率为 ( )
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231804339304658.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231804339304658.png)
A.![]() | B.![]() | C.![]() | D.![]() |
B
设AF2=m,AF1=x,根据双曲线的基本性质及△ABF2是等腰三角形,用m分别表示出x,a,c,进而求得离心率
.
解答:解:设AF2=m,AF1=x
又AB=AF2,则BF1=m-x=2a,BF2=
m.
BF2-BF1=2a,即
m-2a=2a,故a=
m,
又 m-x=2a,解得 x=
m,
在△AF1F2中,由勾股定理知,2c=
=
m
所以双曲线的离心率e=
=
=![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180433977543.png)
故选B.
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180434055332.png)
解答:解:设AF2=m,AF1=x
又AB=AF2,则BF1=m-x=2a,BF2=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180434055344.png)
BF2-BF1=2a,即
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180434055344.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180434102433.png)
又 m-x=2a,解得 x=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180434118477.png)
在△AF1F2中,由勾股定理知,2c=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180434118528.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180434149662.png)
所以双曲线的离心率e=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180434055332.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180434180946.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180433977543.png)
故选B.
![](http://thumb.zyjl.cn/images/loading.gif)
练习册系列答案
相关题目