解:(Ⅰ) 观察知,x=2是圆的一条切线,切点为A
1(2,0),
设O为圆心,根据圆的切线性质,MO⊥A
1A
2,
所以
![](http://thumb.1010pic.com/pic5/latex/127479.png)
,
所以直线A
1A
2的方程为
![](http://thumb.1010pic.com/pic5/latex/127480.png)
.
线A
1A
2与y轴相交于(0,1),依题意知a=2,b=1,
所求椭圆的方程为
![](http://thumb.1010pic.com/pic5/latex/1164.png)
.
(Ⅱ) 椭圆方程为
![](http://thumb.1010pic.com/pic5/latex/1164.png)
,设P(x
0,y
0),A(m,n),B(m,-n),
则有
![](http://thumb.1010pic.com/pic5/latex/127481.png)
,m
2+4n
2-4=0,
在直线AP的方程
![](http://thumb.1010pic.com/pic5/latex/127482.png)
中,令
![](http://thumb.1010pic.com/pic5/latex/127483.png)
,整理得
![](http://thumb.1010pic.com/pic5/latex/127484.png)
.①
同理,
![](http://thumb.1010pic.com/pic5/latex/127485.png)
.②
①×②,并将
![](http://thumb.1010pic.com/pic5/latex/127486.png)
,
![](http://thumb.1010pic.com/pic5/latex/127487.png)
代入得y
Q•y
R=
![](http://thumb.1010pic.com/pic5/latex/127488.png)
=
![](http://thumb.1010pic.com/pic5/latex/127489.png)
=
![](http://thumb.1010pic.com/pic5/latex/127490.png)
=
![](http://thumb.1010pic.com/pic5/latex/127491.png)
.
而
![](http://thumb.1010pic.com/pic5/latex/127492.png)
=
![](http://thumb.1010pic.com/pic5/latex/127493.png)
,
∵|m|<2且m≠0,∴
![](http://thumb.1010pic.com/pic5/latex/127494.png)
,
∴
![](http://thumb.1010pic.com/pic5/latex/127478.png)
.
分析:(Ⅰ)由图易求切点A
1(2,0),根据MO⊥A
1A
2可求直线A
1A
2的方程,从而可求椭圆上顶点,进而得a,b值;
(Ⅱ)设P(x
0,y
0),A(m,n),B(m,-n),则有
![](http://thumb.1010pic.com/pic5/latex/127481.png)
,m
2+4n
2-4=0,写出直线AP方程可求得y
Q,同理求得y
R,于是可得y
Q•y
R,进而得到
![](http://thumb.1010pic.com/pic5/latex/3001.png)
![](http://thumb.1010pic.com/pic5/latex/127495.png)
,再根据m的范围即可求证.
点评:本题考查直线与圆锥曲线的位置关系,考查椭圆的标准方程,考查数形结合思想,考查学生的运算能力、分析问题解决问题的能力,难度较大.