题目内容
双曲线
的左、右顶点分别为
、
,P为其右支上的一点,且
,则
等于( )
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823170505003682.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823170505018334.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823170505159359.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823170505190782.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823170505206566.png)
A.无法确定 | B.![]() | C.![]() | D.![]() |
B
解:设a2=2010,
A1(-a,0),A2(a,0),P(x,y),
kPA1=tan∠PA1A2=
,①
kPA2=-tan∠PA2A1=
,②
由x2-y2=a2得
=1,
①×②,得-tan∠PA1A2tan∠PA2A1=1,
∴tan∠PA1A2tan(5∠PA1A2)=1
即tan(5∠PA1A2)=tan(
-∠PA1A2)
∴5∠PA1A2=
-∠PA1A2
∴∠PA1A2=![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823170505284463.png)
故选B.
A1(-a,0),A2(a,0),P(x,y),
kPA1=tan∠PA1A2=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823170505315491.png)
kPA2=-tan∠PA2A1=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823170505330474.png)
由x2-y2=a2得
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823170505362573.png)
①×②,得-tan∠PA1A2tan∠PA2A1=1,
∴tan∠PA1A2tan(5∠PA1A2)=1
即tan(5∠PA1A2)=tan(
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823170505377421.png)
∴5∠PA1A2=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823170505377421.png)
∴∠PA1A2=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823170505284463.png)
故选B.
![](http://thumb.zyjl.cn/images/loading.gif)
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