顶点在原点,焦点在X轴上且截直线2x-y+1=0所得弦长为√15的抛物线方程

设抛物线为y²=2px
则联立y²=2px和y=2x+1,得：(2x+1)²=2px,即4x²+(4-2p)x+1=0
设两个交点为(x1,y1)(x2,y2)
∴x1+x2=-1+p/2,y1+y2=2(x1+x2)+2=-2+p+2=p
x1x2=1/4,y1y2=(2x1+1)(2x2+1)=4x1x2+2(x1+x2)+1=1-2+p+1=p
弦长的平方=(x1-x2)²+(y1-y2)²
=(x1+x2)²-4x1x2+(y1+y2)²-4y1y2
=1+p²/4-p-1+p²-4p
=5p²/4-5p=15
即p²-4p-12=0,即(p-6)(p+2)=0
则p=6或者p=-2
∴抛物线为y²=12x或者y²=-4x