设抛物线为y^2=2px,其焦点为（p/2,0),过焦点直线的斜率为tan135°=-1其方程为y=-(x-p/2),即：x+y=p/2,设它与抛物线交于(x1,y1),(x2,y2)把直线方程代入抛物线方程可得：x^2-3px+p^2/4=0①,y^2+2py-p^2=0②显然x1、x2...