y^2=2px与直线方程x=ty+r联立消去x得
ay^2+by+c=0
韦达定理得y1y2=-c/a
∴y1^2=2px1 y2^2=2px2
则x1x2=[y1^2/(2p)] [y2^2/(2p)]=(y1y2)^2/(4p^2).
若消y同理 ：y1y2=-√(4p²x1x2)