把y=x-1代人y²=2px中得
(x-1)²=2px,即:x²-(2+2p)x+1=0
∴x1+x2=2+2p x1x2=1
∵|AB|²=(1+k²)(x1-x2)²=(1+1²)[(x1+x2)²-4x1x2]=2[(2+2p)²-4]=8p²+16p
∴64=8p²+16p
解得:p=-4或者p=2
∴方程为:y²=-8x,或者y²=4x
把y=x-1代人y²=2px中得
(x-1)²=2px,即:x²-(2+2p)x+1=0
∴x1+x2=2+2p x1x2=1
∵|AB|²=(1+k²)(x1-x2)²=(1+1²)[(x1+x2)²-4x1x2]=2[(2+2p)²-4]=8p²+16p
∴64=8p²+16p
解得:p=-4或者p=2
∴方程为:y²=-8x,或者y²=4x