Question

已知直线y=2x+1被抛物y的平方=2px线截得的弦长根号15，求抛物线的方程。

快点

Answer 1

设A(x1，y1)，B(x2，y2)
将y=2x+1代入y^2=2px得4x^2+(4-2p)x+1=0
x1+x2=(p-2)/2，x1*x2=1/4
弦长公式：|AB|^2=(1+k^2)* |x1-x2|=5[(x1+x2)^2-4x1*x2]=15
解得p=6或p=-2
∴抛物线方程为y^2=12x或y^2=-4x

Answer 2

将y=2x+1代入y^2=2px得y^2-px+p=0
yi+y2=p，y1y2=p
(y1-y2)^2=(y1+y2)^2-4y1y2=p^2-p
(y1-y2)/(x1-x2)=2
|AB|^2=(x1-x2)^2+(y1-y2)^2=5/4*(y1-y2)^2=5/4*(p^2-4p)=15
得p=6或-2
y^2=12x或y^2=-4x