设抛物线方程为x²=2py
将直线y=2x-1代入
x²-4px+2p=0
由韦达定理x1+x2=4px1*x2=2p
则(x1-x2)²=(x1+x2)²-4x1*x2=16p²-8p
又y1-y2=(2x1-1)-(2x2-1)=2(x1-x2)
所以弦长=√[(x1-x2)²+(y1-y2)²]
=√[5(x1-x2)²]
=√[5(16p²-8p)]=2√10
两边平方 5(16p²-8p)=40
2p²-p-1=0
(2p+1)(p-1)=0
解得p=-1/2或1
所以抛物线方程为x²=-y或x²=2y