弦AB过抛物线y^2=2px的焦点,求弦AB中点M的轨迹方程

焦点(p/2,0)
y=k(x-p/2)
则k²(x²-px+p²/4)=2px
k²x²-(k²p+2p)x+p²/4=0
x1+x2=(k²p+2p)/k²
y1+y2=k(x1-p/2)+k(x2-p/2)
=k(x1+x2)-kp
=2p/k
x=(x1+x2)/2,y=(y1+y2)/2
y/x=2pk/(k²p+2p)=2k/(k²+1)
y(k²+1)=2kx
y=k(x-p/2)
k=y/(x-p/2)
所以y*[y²/(x²-pk+p²/4)+1]=2xy/(x-p/2)
y²+x²-pk+p²/4=2x(x-p/2)
y²+p²/4=x²