试题分析:(Ⅰ)先判断RQ是线段FP的垂直平分线,从而可得动点Q的轨迹C是以F为焦点,l为准线的抛物线;
(Ⅱ)设M(m,-p),两切点为A(x
1,y
1),B(x
2,y
2),求出切线方程,从而可得x
1,x
2为方程x
2-2mx-4p
2=0的两根,进一步可得直线AB的方程,即可得到直线恒过定点(0,p);
解:(1)依题意知,点
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000516207299.png)
是线段
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000516238364.png)
的中点,且
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000516254418.png)
⊥
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000516238364.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240005161762591.png)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000516254418.png)
是线段
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000516238364.png)
的垂直平分线. ∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000516348602.png)
.
故动点
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000515802328.png)
的轨迹
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000515817306.png)
是以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000516488303.png)
为焦点,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000515833272.png)
为准线的抛物线,
其方程为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000516192828.png)
.
(2)设
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000516535738.png)
,两切点为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000516769612.png)
,
∴两条切线方程为x
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000516816195.png)
x=2p(y+y
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000516816195.png)
) ①
x
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000516847240.png)
x=2p(y+y
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000516847240.png)
) ②
对于方程①,代入点
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000516535738.png)
, 又
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000516894723.png)
, 整理得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000516925827.png)
, 同理对方程②有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000516940838.png)
, 即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000517174424.png)
为方程
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000517190796.png)
的两根.
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000517206904.png)
③
设直线
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000516160392.png)
的斜率为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000517252313.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240005172681529.png)
所以直线
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000516160392.png)
的方程为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240005173151161.png)
,展开得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240005173301047.png)
,代入③得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000517346777.png)
, ∴直线恒过定点
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000517377483.png)
.
点评:解决该试题的关键是正确运用圆锥曲线的定义和韦达定理,来表示根与系数的关系的运用。