试题分析:(1)设出P、Q的坐标,求得向量的坐标,利用
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235504873675.png)
,P(x
0,y
0)在双曲线上,即可求得结论;
(2)利用三点共线建立方程,利用P(x
0,y
0)在双曲线上,即可求得轨迹方程;
(3)用坐标表示
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232355051541222.png)
,利用韦达定理,求得模长,从而可得函数关系式,进而可求其范围.
解:(1)由题,得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235505201815.png)
,设
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235505216879.png)
则
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232355052481362.png)
由
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232355052631400.png)
……①
又
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235505294640.png)
在双曲线上,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235505310711.png)
……②
联立①、②,解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235505341447.png)
由题意,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235505357624.png)
∴点T的坐标为(2,0)
(2)设直线A
1P与直线A
2Q的交点M的坐标为(x,y)
由A
1、P、M三点共线,得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235505497893.png)
……③
由A
2、Q、M三点共线,得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235505513884.png)
……④ 联立③、④,解得
∵
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235505294640.png)
在双曲线上,∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235505684949.png)
∴轨迹E的方程为
(3)容易验证直线l的斜率不为0。
故可设直线l的方程为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232355057471014.png)
中,得
设
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232355058871098.png)
则由根与系数的关系,得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235505903758.png)
……⑤
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235506090702.png)
……⑥
∵
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235506106645.png)
∴有
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235506121691.png)
将⑤式平方除以⑥式,得
由
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232355063081608.png)
∵
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232355063401972.png)
又
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232355063712236.png)
故
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232355050921224.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232355064022388.png)
考点:
点评:解决该试题的关键是借助于向量关系式来表示得到坐标,同时能利用三点共线,进而得到坐标关系,解得轨迹方程。易错点就是设而不求的思想,在运算中的准确表示。