设抛物线方程为y2=2px,和直线两交点为A(x1,y1),B(x2,y2),k=kAB=2
y=2x+1代入y2=2px得4x2+(4-2p)x+1=0,∴x1+x2=(p/2)-1,x1x2=1/4
弦长|AB|2=15=(1+k2)(x12-x22)=(1+k2){(x1+x2)2-4x1x2}=(1+22){((p/2)-1)2-4·(1/4)}解得p=-2或6