题目内容
已知双曲线
的左右焦点分别 为F1、F2,P是准线上一点,且
·
=0,
·
=4ab,则双曲线的离心率是
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231805241941091.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180524209411.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180524225448.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180524240434.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180524256465.png)
A.![]() | B.![]() | C.2 | D.3 |
B
分析:设右准线与x轴的交点为A,根据PF1⊥PF2,利用射影定理可得|PA|2=|AF1|×|AF2|,利用P到x轴的距离为
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180524396521.png)
解:∵P是右准线上一点,P到x轴的距离为
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180524396521.png)
∴可设P(
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180524428420.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180524396521.png)
设右准线与x轴的交点为A,
∵PF1⊥PF2,
∴|PA|2=|AF1|×|AF2|
∴(
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180524396521.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180524428420.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180524428420.png)
∴4a2b2=(c2-a2)(c2+a2)
∴4a2=c2+a2
∴3a2=c2
∴e=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180524599352.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823180524615344.png)
故选B.
![](http://thumb.zyjl.cn/images/loading.gif)
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