Question

已知过抛物线y^2=2px(p>0)的焦点 斜率为2根号2的直线交抛物线于A(x1,y1),B(x2,y2) -(x1<x2)-两点且|AB|=9

求该抛物线的方程

Answer 1

|AB|=x1+p/2+x2+p/2=x1+x2+p (x1+x2)=9-p
|AB|=√(k^2+1)|x1-x2| =3|x1-x2| =9
(x1-x2)^2=9
y=k(x-p/2)
k^2(x^2-px+p^2/4)=2px
k^2x^2-(k^2p+2p)x+k^2p^2/4=0
x1x2=p^2/4
(x1+x2)^2=(9-p)^2=(x1-x2)^2+4x1x2=9+p^2
81-18p+p^2=9+p^2
72-18p=0
p=4
方程为y^2=8x