顶点在原点,焦点在x轴上的抛物线与直线y=2x+1相交于A,B两点,且|AB|=根号15,求抛物线方程

y^2=2px
把直线带入
4x^2+4x+1=2px
4x^2+(4-2p)x+1=0
x1+x2=(2p-4)/4=(p-2)/2,x1x2=1/4
(x1-x2)^2=(x1+x2)^2-4x1x2=(p-2)^2/4-1=(p^2-4p)/4
y=2x+1
所以(y1-y2)^2=[(2x1+1)-(2x2+1)]^2=[2(x1-x2)]^2=4*(x1-x2)^2
所以AB^2=(x1-x2)^2+(y1-y2)^2=5(x1-x2)^2=15
(p^2-4p)/4=3
p^2-4p-12=0
(p-6)(p+2)=0
p=6,p=-2
y^2=12x,y^2=-4x