过抛物线y^2=2px（p>0）焦点坐标F（p/2,0)设直线斜率k：y=k(x-p/2),代入y^2=2px：[k(x-p/2)]^2=2pxk^2x^2 - (k^2p+2p)x + k^2p^2/4 = 0根据韦达定理：x1x2 = (k^2p^2/4)/k^2 = p^2/4 = 定值,得证.要算y1y2 就把y=k(x...