已知，抛物线C:y^2=-4x.故可设点A(-a^2,2a),B(-b^2,2b).M(m,0).由题设知，点A,B,(-2,0)共线，===》ab=-2.再由题设知，[-2a/(m+a^2)]+[-2b/(m+b^2]=0.===>(m+ab)(a+b)=0.===>m+ab=0.===>m=-ab=2.===>M(2,0).
)假设存在M符合题意
设A(x1,y1)B(x2,y2) (y1>0,y2<0)
(x1<0,x2<0)
由于：X轴平分角AMB
则由角平分线性质可得：
B(x2,y2)关于X轴的对称点B'(x2,-y2)在直线AM上
即:直线AB'与X轴交于定点M
则：kAB'=[y1-(-y2)]/(x1-x2)=(y1+y2)/(x1-x2)
由于A,B在抛物线C上
则有：y1^2=-4x1,y2^2=-4x2
两式相减得：(y1+y2)(y1-y2)=-4(x1-x2)
则：(y1+y2)/(x1-x2)=-4/(y1-y2)
即：kAB'=-4/(y1-y2)
则：AB':y-y1=[-4/(y1-y2)](x-x1)
y=k(x+2)与抛物线C联立得：
k^2x^2+(4k^2+4)x+4k^2=0
则：x1+x2=-(4k^2+4)/k^2,
x1x2=4k^2/k^2=4
由于AB':y-y1=[-4/(y1-y2)](x-x1)与X轴交于M
则令y=0
则：-y1=[-4/(y1-y2)](xM-x1)
即：xM=(y1^2-y1y2)/4
+x1
=(-4x1-y1y2)/4
+(4x1)/4
=(-y1y2)/4
由于：(y1^2)*(y2^2)=(-4x1)*(-4x2)=16x1x2=64
且y1y2<0
则：-y1y2=8
则：xM=(-y1y2)/4=2
即定点M(2,0)
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