设直线AB方程为x=my+n,与抛物线y²=2px联立消去x得：
y²-2pmy-2pn=0.
设A(x1,y1),B(x2,y2).则 y1+y2=2pm,y1y2=-2pn.
x1+x2=m(y1+y2)+2n=2pm^2+2n
x1x2=(my1+n)(my2+n)=m^2*y1y2+mn(y1+y2)+n^2=n^2
因为向量MA*向量MB=0,所以(x1-x0)(x2-x0)+(y1-y0)(y2-y0)=0.
即x1x2-x0(x1+x2)+x0^2+y1y2-y0(y1+y2)+y0^2=0
n^2-x0(2pm^2+2n)+x0^2-2pn-2pmy0+y0^2=0 .
n^2-y0*m^2-ny0^2/p-2pn-2pmy0+y0^2+y0^4/(4p^2)=0
不知道是你自己修改了题目还是抄错了,原题M是原点,改成这个算起来就像上面一样,难以化简
如果是原点,按照上面写一遍就OK了