解:(Ⅰ)設(shè)曲線C上任意一點(diǎn)P(x,y),又F(1,0),N(-1,y),
從而
![](http://thumb.1010pic.com/pic5/latex/27296.png)
,
![](http://thumb.1010pic.com/pic5/latex/27297.png)
,
則
![](http://thumb.1010pic.com/pic5/latex/27298.png)
=
![](http://thumb.1010pic.com/pic5/latex/27299.png)
,
由
![](http://thumb.1010pic.com/pic5/latex/27295.png)
,得
![](http://thumb.1010pic.com/pic5/latex/27300.png)
,
即
![](http://thumb.1010pic.com/pic5/latex/27301.png)
.
化簡得y
2=4x,即為所求的P點(diǎn)的軌跡C的對應(yīng)的方程.
(Ⅱ)設(shè)A(x
1,y
1)、B(x
2,y
2),
MA:y=k
1(x-1)+2,
MB:y=k
2(x-1)+2.
將y=k
1(x-1)+2與y
2=4x聯(lián)立,得:
![](http://thumb.1010pic.com/pic5/latex/27302.png)
由
![](http://thumb.1010pic.com/pic5/latex/27303.png)
,得
![](http://thumb.1010pic.com/pic5/latex/27304.png)
①
同理
![](http://thumb.1010pic.com/pic5/latex/27305.png)
②
而AB直線方程為:
![](http://thumb.1010pic.com/pic5/latex/27306.png)
,
即
![](http://thumb.1010pic.com/pic5/latex/27307.png)
③
由①②:y
1+y
2=
![](http://thumb.1010pic.com/pic5/latex/27308.png)
![](http://thumb.1010pic.com/pic5/latex/27309.png)
代入③得,
![](http://thumb.1010pic.com/pic5/latex/27310.png)
,
整理得k
1k
2(x+y+1)+6+y=0.
則
![](http://thumb.1010pic.com/pic5/latex/27311.png)
,故直線AB經(jīng)過定點(diǎn)(5,-6).
分析:(Ⅰ)設(shè)出動點(diǎn)P的坐標(biāo),求出N點(diǎn)的坐標(biāo),再求出向量
![](http://thumb.1010pic.com/pic5/latex/27312.png)
,然后代入
![](http://thumb.1010pic.com/pic5/latex/27295.png)
整理即可得到點(diǎn)P的軌跡C的方程;
(Ⅱ)設(shè)出點(diǎn)A,B的坐標(biāo),寫出直線MA,MB的方程,和拋物線聯(lián)立后利用根與系數(shù)關(guān)系求出A點(diǎn)和B點(diǎn)的縱坐標(biāo),然后求出兩縱坐標(biāo)的和與積,然后由直線方程的兩點(diǎn)式寫出AB的直線方程,把兩縱坐標(biāo)的和與積代入直線方程后,利用直線系方程的知識可求出直線AB經(jīng)過的定點(diǎn).
點(diǎn)評:本題考查了拋物線的方程,考查了直線與拋物線的綜合,訓(xùn)練了一元二次方程的根與系數(shù)關(guān)系,考查了直線系方程,此題是有一定難度題目.