题目内容：

过抛物线y2;=2px的顶点作互相垂直的弦OA,OB,过O作OM垂直于AB,垂足为M,求M点的轨迹方程

优质解答

设kOA=k kOB=-1/k则A(2P/k^2,2P/k) B(2Pk^2,-2Pk)kAB=k/(1-k^2)AB：y+2Pk=[k/(1-k^2)](x-2Pk^2)即y=[k/(1-k^2)](x-2P)∴AB经过定点(2P,0)AB：y+2Pk=[k/(1-k^2)](x-2Pk^2)①OM：y=[-(1-k^2)/k]x② ==>k^2x=x+ky③两式...