本试题主要是考查了双曲线方程的求解,以及直线与双曲线的位置关系的综合运用。结合韦达定理和弦长公式,以及向量的坐标关系式,得到参数的求解。
(1)根据双曲线的定义可以得到双曲线的方程的求解。
(2)联立方程组,得到相交弦的长度以及韦达定理得到直线的方程。
(3)根据
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632050791.png)
,得到坐标关系式,结合上一问的结论,可知参数m的等式,得到结论。
解:(Ⅰ)由双曲线的定义可知,曲线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222631769318.png)
是以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632097897.png)
为焦点的双曲线的左支,且
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632112549.png)
,易知
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632128352.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232226321598625.png)
故曲线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222631769318.png)
的方程为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222631925772.png)
……….4分
(Ⅱ) 设
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632222922.png)
,由题意建立方程组
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632237849.png)
消去
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632315310.png)
,得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632331924.png)
又已知直线与双曲线左支交于两点
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632346423.png)
,有
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232226323622259.png)
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632409551.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632424169.png)
……….6分
又∵
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632440894.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232226324871044.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232226325021222.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232226325181246.png)
依题意得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232226325341329.png)
整理后得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632596805.png)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632612532.png)
或
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632627526.png)
但
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632409551.png)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632674584.png)
故直线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632690396.png)
的方程为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222631941772.png)
……….9分
(Ⅲ)设
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632736657.png)
,由已知
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632050791.png)
,得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232226327681113.png)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232226328141383.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632830575.png)
又
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632846940.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232226328611458.png)
∴点
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632892964.png)
将点
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222632924313.png)
的坐标代入曲线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222631769318.png)
的方程,得
得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222633095444.png)
,但当
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222633111428.png)
时,所得的点在双曲线的右支上,不合题意 ∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222631956424.png)
,…13分