(Ⅰ)解:依题意知,点R是线段FP的中点,且RQ⊥FP,
∴RQ是线段FP的垂直平分线.---------------------------------------(2分)
∴|PQ|=|QF|.
∴动点Q的轨迹C是以F为焦点,l为准线的抛物线,其方程为:x
2=4py(p>0).--------------------(4分)
(Ⅱ)证明:设M(m,-p),两切点为A(x
1,y
1),B(x
2,y
2)
由x
2=4py得
![](http://thumb.1010pic.com/pic5/latex/85740.png)
,求导得y′=
![](http://thumb.1010pic.com/pic5/latex/85741.png)
.
∴两条切线方程为
![](http://thumb.1010pic.com/pic5/latex/85742.png)
①
![](http://thumb.1010pic.com/pic5/latex/85743.png)
②-------------------(6分)
对于方程①,代入点M(m,-p)得,
![](http://thumb.1010pic.com/pic5/latex/85744.png)
,
又
![](http://thumb.1010pic.com/pic5/latex/85745.png)
∴
![](http://thumb.1010pic.com/pic5/latex/85746.png)
整理得:
![](http://thumb.1010pic.com/pic5/latex/85747.png)
同理对方程②有
![](http://thumb.1010pic.com/pic5/latex/85748.png)
即x
1,x
2为方程x
2-2mx-4p
2=0的两根.
∴x
1+x
2=2m,x
1x
2=-4p
2 ③-----------------------(8分)
设直线AB的斜率为k,
![](http://thumb.1010pic.com/pic5/latex/85749.png)
=
![](http://thumb.1010pic.com/pic5/latex/85750.png)
所以直线AB的方程为
![](http://thumb.1010pic.com/pic5/latex/85751.png)
,展开得:
![](http://thumb.1010pic.com/pic5/latex/85752.png)
,
代入③得:
![](http://thumb.1010pic.com/pic5/latex/85753.png)
∴直线恒过定点(0,p).-------------------------------------(10分)
(Ⅲ) 证明:由(Ⅱ)的结论,设M(m,-p),A(x
1,y
1),B(x
2,y
2)
且有x
1+x
2=2m,x
1x
2=-4p
2,
∴k
MA=
![](http://thumb.1010pic.com/pic5/latex/85754.png)
,k
MB=
![](http://thumb.1010pic.com/pic5/latex/85755.png)
----------------------------(11分)
∴
![](http://thumb.1010pic.com/pic5/latex/85756.png)
=
![](http://thumb.1010pic.com/pic5/latex/85757.png)
=
![](http://thumb.1010pic.com/pic5/latex/85758.png)
=-
![](http://thumb.1010pic.com/pic5/latex/85759.png)
------(13分)
又∵
![](http://thumb.1010pic.com/pic5/latex/85760.png)
=-
![](http://thumb.1010pic.com/pic5/latex/85761.png)
,
∴
![](http://thumb.1010pic.com/pic5/latex/85762.png)
即直线MA,MF,MB的斜率的倒数成等差数列.----------------------------(14分)
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);
(Ⅲ) 由(Ⅱ)的结论,设M(m,-p),A(x
1,y
1),B(x
2,y
2),且有x
1+x
2=2m,x
1x
2=-4p
2,从而可得k
MA=
![](http://thumb.1010pic.com/pic5/latex/85754.png)
,k
MB=
![](http://thumb.1010pic.com/pic5/latex/85755.png)
,由此可证直线MA,MF,MB的斜率的倒数成等差数列.
点评:本题考查抛物线的定义,考查直线恒过定点,考查直线的向量,解题的关键是正确运用韦达定理,属于中档题.