试题分析:(1)因为直线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528227280.png)
过右焦点
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528632511.png)
,斜率为1,
所以直线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528227280.png)
的方程为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528663477.png)
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528695510.png)
.
坐标原点
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528305292.png)
到直线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528227280.png)
的距离为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528757413.png)
,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528773639.png)
,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528804304.png)
. …2分
因为离心率为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528757413.png)
,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528851458.png)
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528866556.png)
,
所以椭圆C的方程为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528554654.png)
. …4分
(2)因为直线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528227280.png)
过右焦点,所以当直线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528227280.png)
斜率不存在时,直线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528227280.png)
方程为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528991355.png)
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235529022926.png)
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235529038861.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528429289.png)
为右端点时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235529085742.png)
,
所以此时没有符合要求的点
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528429289.png)
.
当直线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528227280.png)
斜率存在时,设直线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528227280.png)
方程为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235529163610.png)
,
由
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232355291941093.png)
得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232355292091082.png)
. …7分
设点
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235529225423.png)
的坐标分别为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235529241544.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235529272567.png)
,
则
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235529287849.png)
,因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235529506660.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235529521665.png)
,
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235529553855.png)
,
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232355295682241.png)
,
所以点
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528429289.png)
的坐标为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235529615954.png)
,且符合椭圆方程,
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232355296461341.png)
,解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235529662608.png)
所以点
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528429289.png)
的坐标为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528585756.png)
或
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235528601767.png)
. …12分
点评:设直线方程时要注意斜率存在与不存在两种情况,求解直线与椭圆位置关系问题时,通常要联立方程组,运算量比较大,应该仔细计算,并且要注意通性通法的应用,加强解题的规范性.