试题分析:(Ⅰ)根据椭圆E:椭圆E:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235058908691.png)
=1(a>b>o)的离心率e=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235058923411.png)
,可得a
2=2b
2,利用椭圆E:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235058908691.png)
=1经过点(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235058939334.png)
,1)我们有
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235059282644.png)
,从而可求椭圆E的标准方程;
(Ⅱ)连接OM,OP,OQ,设M(-4,m),由圆的切线性质及∠PMQ=60°,可知△OPM为直角三角形且∠OMP=30°,从而可求M(-4,4),进而以OM为直径的圆K的方程为(x+2)
2+(y-2)
2=8与圆O:x
2+y
2=8联立,两式相减可得直线PQ的方程.
解:(1)椭圆的标准方程为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235058986712.png)
﹍﹍﹍﹍﹍﹍﹍4分
(2)连接QM,OP,OQ,PQ和MO交于点A,
有题意可得M(-4,m),∵∠PMQ=60
0![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235059329834.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232350593444677.jpg)
∴∠OMP=30
0,∵
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232350593761575.png)
,
∵m>0,∴m=4,∴M(-4,4) ﹍﹍﹍﹍﹍﹍﹍7分
∴直线OM的斜率
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235059391558.png)
,有MP=MQ,OP=OQ可知OM⊥PQ,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235059422537.png)
,设直线PQ的方程为y=x+n ﹍﹍﹍﹍﹍﹍﹍9分
∵∠OMP=30
0,∴∠POM=60
0,∴∠OPA=30
0,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235059438770.png)
,即O到直线PQ的距离为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235059469336.png)
, ﹍﹍﹍﹍10分
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823235059485749.png)
(负数舍去),∴PQ的方程为x-y+2=0. ﹍﹍﹍﹍12分
点评:解题的关键是确定M的坐标,进而确定以OM为直径的圆K的方程.