【答案】
分析:(I)先利用离心率条件求出a,c的关系式,再利用右准线方程得到a,c的另一个关系式结合a,b,c的关系即可求得a,b.最后写出椭圆的方程即可;
(II)先圆心到直线的距离等于半径可得t和k满足的关系式,把直线l的方程与椭圆方程联立求出A、B两点的坐标,再利用
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125232308696848/SYS201310251252323086968020_DA/0.png)
即可求出m与k的关系式;
(III)用类似于(2)的方法求出m,k之间的关系式,求出弦AB的长,再把△AOB面积整理成关于m的函数;利用函数的单调性求出△AOB面积的取值范围即可.
解答:解:(Ⅰ)由条件知:
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125232308696848/SYS201310251252323086968020_DA/1.png)
.
得
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125232308696848/SYS201310251252323086968020_DA/2.png)
,
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125232308696848/SYS201310251252323086968020_DA/3.png)
.b=1.
∴椭圆C的方程为:
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125232308696848/SYS201310251252323086968020_DA/4.png)
.(3分)
(Ⅱ)依条件有:
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125232308696848/SYS201310251252323086968020_DA/5.png)
,即t
2=2(1+k
2).(4分)
由
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125232308696848/SYS201310251252323086968020_DA/6.png)
得:(3k
2+1)x
2+6ktx+3t
2-3=0.△=12(k
2-1),设A(x
1,y
1),B(x
2,y
2),则
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125232308696848/SYS201310251252323086968020_DA/7.png)
,
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125232308696848/SYS201310251252323086968020_DA/8.png)
又t
2=2k
2+1,∴
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125232308696848/SYS201310251252323086968020_DA/9.png)
=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125232308696848/SYS201310251252323086968020_DA/10.png)
由
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125232308696848/SYS201310251252323086968020_DA/11.png)
在
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125232308696848/SYS201310251252323086968020_DA/12.png)
方向上的投影是p,得
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125232308696848/SYS201310251252323086968020_DA/13.png)
(7分)∴
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125232308696848/SYS201310251252323086968020_DA/14.png)
(10分)
(Ⅲ)由弦长公式得
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125232308696848/SYS201310251252323086968020_DA/15.png)
.
由
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125232308696848/SYS201310251252323086968020_DA/16.png)
,得
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125232308696848/SYS201310251252323086968020_DA/17.png)
∴
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125232308696848/SYS201310251252323086968020_DA/18.png)
(12分)∴
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125232308696848/SYS201310251252323086968020_DA/19.png)
.
又
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125232308696848/SYS201310251252323086968020_DA/20.png)
,∴
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125232308696848/SYS201310251252323086968020_DA/21.png)
.(14分)
点评:本题是对函数,向量,抛物线以及圆的综合考查、函数单调性的应用等基础知识,考查运算求解能力,考查数形结合思想、化归与转化思想.