由x-y-1=0得y=x-1,代入y^2=2px得
x²-（2＋2p）x＋1＝0
设A(x1,y1),B(x2,y2)
则x1＋x2＝2＋2p,x1x2＝1
且y1=x1-1,y2=x2-1
∴|AB|=根号下[(x2-x1)²+(y2-y1)²]
=根号下[2(x2-x1)²]
=根号下[2(x2+x1)²-8x1x2]
=根号下[2(2+2p)²-8]
∴2(2+2p)²-8=64
解得p=2或p=-4
又△>0 即(2+2p)²-4>0
∴p=2或p=-4均成立
∴抛物线方程为y²=4x或y²=-8x