解
设A,B,P三点坐标分别为A(x1,y1),B(x2,y2),P(x,y)
代入椭圆方程x^2/2+y^2=1,得
x1^2/2+y1^2=1
x2^2/2+y2^2=1
两式相减得:(x1-x2)(x1+x2)+(y1-y2)(y1+y2)=0
又:x1+x2=2x,y1+y2=2y,即(x1-x2)x+(y1-y2)y=0
∴直线AB的斜率k=(y1-y2)/(x1-x2)=-x/y
右焦点F的坐标为(1,0)
又k=(y-0)/(x-1)
∴y/(x-1)=-x/y
∴AB中点M的轨迹 y^2=-x(x-1)