设OA：y=kx ；OB：y=-x/k 由y=kx;y^2=4x得x=4/k^2,y=4/k,即A(4/k^2,4/k) 同理 B(4k^2,-4k) 则xP=1/2(xA+xB)=2(k^2+1/k^2)≥4 yP=1/2(yA+yB)=2(1/k-k) ∴xP=2[(1/k-k)^2+2]=2[(yP/2)^2+2] ∴xP/2=(yP/2)^2+2 故P点轨迹为x/2=y^2/4+2 即y^2=2x-8