21. (本题满分14分)
解:(Ⅰ)点C的轨迹方程为
,.............................4'
(Ⅱ).由方程组
|
y=k(x+1)
ky2+y-k=0.
设A(x1,y1)、B(x2,y2),由韦达定理
.............6'
∵A、B在抛物线y2=-x上,
∴y12=-x1,y22=-x2,y12·y22=x1x2.
设直线l与x轴交于点N,则N(-1,0)
∵S△OAB=S△OAN+S△OBN
=
|ON||y1|+
|ON||y2|
=
|ON|·|y1-y2|,
∴S△OAB=
·1·![]()
=![]()
.............................................8'
∵S△OAB=
,
∴
=![]()
.解得k=±
...................................10'
(Ⅲ)![]()
![]()
故存在唯一的合乎题意的点M(0,0).............................14'