试题分析:(Ⅰ)设Q(x,y),P(x
0,y
0),则D(x
0,0),由Q为线段PD的中点,知x
0=x,y
0=2y,由P(x
0,y
0)在圆x
2+y
2=16上,知x
02+y
02=16,由此能求出点Q的轨迹方程.
(Ⅱ)设直线AB的方程为y-1=k(x-1).由y=k(x-1)+1,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000501512677.png)
,得(1+4k
2)x+8k(1-k)x+4(1-k)
2-16=0,设A(x
1,y
1),B(x
2,y
2),x
1+x
2=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000501528776.png)
,而M(1,1)是AB中点,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000501543544.png)
=1,由此能求出直线方程.
(1)设Q(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000501559432.png)
) P(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000501574465.png)
) 则D(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000501590423.png)
)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000501606712.png)
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000501621629.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000501715790.png)
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000501481719.png)
为所求。 …………4分
(2)法1:依题意显然
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000501762363.png)
的斜率存在,设直线AB的斜率为k,则AB的方程可设为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000501777658.png)
。
由
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240005017931267.png)
得
得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240005018401295.png)
…………7分
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240005018551700.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240005018711200.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240005019021249.png)
…………10分
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240005019181626.png)
…………12分
法2:(直接求k):设A(x
1,y
1),B(x
2,y
2)。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240005019332131.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240005019641277.png)
…………6分
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240005019641631.png)
…………8分
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240005019961712.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240005020111422.png)
…………10分
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240005020271613.png)
…………12分
点评:解决该试题的关键是体现了解析几何中设而不求的解题思想,联立方程组,,转化为二次方程的根的问题,结合韦达定理得到。