(I)由e和橢圓過點
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232210830624.png)
可得到關于a,b的兩個方程,從而解出a,b值求出橢圓的方程.
(II) 設
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232210518280.png)
的方程為
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232211376624.png)
,由已知
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232211423410.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232211485262.png)
得:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408232322115011556.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408232322116261064.png)
=0,
然后直線方程與橢圓方程聯(lián)立消y后得到關于x的一元二次方程,利用韋達定理建立關于k的方程求出k值.
(III)要討論AB斜率存在與不存在兩種情況.研究當AB斜率存在時,由已知
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232211423410.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232211485262.png)
,得
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232211751797.png)
,又
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232210643616.png)
在橢圓上,所以
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408232322117971259.png)
,從而證明出
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408232322118131136.png)
為定值.
解:(Ⅰ)∵
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408232322118911564.png)
……2分
∴
∴橢圓的方程為
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232211095643.png)
……………3分
(Ⅱ)依題意,設
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232210518280.png)
的方程為
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232211376624.png)
由
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408232322120311869.png)
顯然
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232212047426.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408232322120631130.png)
………………5分
由已知
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232211423410.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232211485262.png)
得:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408232322115011556.png)
解得
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232211314480.png)
……………………6分
(Ⅲ)①當直線
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232212219390.png)
斜率不存在時,即
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232212234620.png)
,
由已知
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232211423410.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232211485262.png)
,得
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232211751797.png)
又
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232210643616.png)
在橢圓上,
所以
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408232322124371276.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408232322118131136.png)
,三角形的面積為定值.………7分
②當直線
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232212219390.png)
斜率存在時:設
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232212219390.png)
的方程為
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232212515586.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408232322125311850.png)
必須
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232212047426.png)
即
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232212577922.png)
得到
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232212593834.png)
,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232212609759.png)
………………9分
∵
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232210799436.png)
,∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408232322126871398.png)
代入整理得:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823232212733615.png)
…………………10分
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408232322127491723.png)
…………11分
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408232322127801297.png)
所以三角形的面積為定值. ……12分