由题可知F( p 2 ,0),则该直线方程为:y=x? p 2 代入y2=2px(p>0)得:x2?3px+ p2 4 =0,设M(x1,y1),N(x2,y2),则有x1+x2=3p,∵|MN|=8,∴x1+x2+p=8,即3p+p=8,解得p=2∴抛物线的方程为:y2=4x.