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)