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

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)①
OM：y=[-(1-k^2)/k]x② ==>k^2x=x+ky③
两式相乘 y(y+2Pk)=-x(x-2Pk^2)
即x^2+y^2-2Pk^2x+2Pky=0
代人③ 得x^2+y^2-2Px=0 即(x-P)^2+y^2=P^2 (x≠0)