【答案】
分析:(Ⅰ)先利用離心率得到一個關于參數的方程,再利用x軸被曲線C
2:y=x
2-b截得的線段長等于C
1的長半軸長得另一個方程,兩個方程聯(lián)立即可求出參數進而求出C
1,C
2的方程;
(Ⅱ)(i)把直線l的方程與拋物線方程聯(lián)立可得關于點A、B坐標的等量關系,再代入求出k
MA•k
MB=-1,即可證明:MD⊥ME;
(ii)先把直線MA的方程與拋物線方程聯(lián)立可得點A的坐標,再利用弦長公式求出|MA|,同樣的方法求出|MB|進而求出S
1,同理可求S
2.再代入已知就可知道是否存在直線l滿足題中條件了.
解答:解:(Ⅰ)由題得e=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/0.png)
,從而a=2b,又2
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/1.png)
=a,解得a=2,b=1,
故C
1,C
2的方程分別為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/2.png)
,y=x
2-1.
(Ⅱ)(i)由題得,直線l的斜率存在,設為k,則直線l的方程為y=kx,
由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/3.png)
得x
2-kx-1=0.
設A(x
1,y
1),B(x
2,y
2),則x
1,x
2是上述方程的兩個實根,
于是x
1+x
2=k,x
1x
2=-1,又點M的坐標為(0,-1),
所以k
MA•k
MB=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/4.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/5.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/6.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/7.png)
=-1.
故MA⊥MB,即MD⊥ME.
(ii)設直線MA的斜率為k
1,則直線MA的方程為y=k
1x-1.
由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/8.png)
,解得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/9.png)
或
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/10.png)
.
則點A的坐標為(k
1,k
12-1).
又直線MB的斜率為-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/11.png)
,同理可得點B的坐標為(-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/12.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/13.png)
-1).
于是s
1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/14.png)
|MA|•|MB|=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/15.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/16.png)
•|k
1|•
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/17.png)
•|-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/18.png)
|=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/19.png)
.
由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/20.png)
得(1+4k
12)x
2-8k
1x=0.
解得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/21.png)
或,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/22.png)
,則點D的坐標為(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/23.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/24.png)
).
又直線ME的斜率為-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/25.png)
.同理可得點E的坐標為(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/26.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/27.png)
).
于是s
2=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/28.png)
|MD|•|ME|=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/29.png)
.
故
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/30.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/31.png)
,解得k
12=4或k
12=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/32.png)
.
又由點A,B的坐標得,k=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/33.png)
=k
1-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/34.png)
.所以k=±
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/35.png)
.
故滿足條件的直線存在,且有兩條,其方程為y=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/36.png)
x和y=-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103759805282794/SYS201311031037598052827020_DA/37.png)
x.
點評:本題是對橢圓與拋物線以及直線與拋物線和直線與橢圓的綜合問題的考查.是一道整理過程很麻煩的題,需要要認真,細致的態(tài)度才能把題目作好.