Question

已知抛物线C：y2=4x的焦点为F，过点F的直线l与C相交抛物线于A，B两点．若AB＝8 求直线l的方程。

Answer 1

F(1,0)
y=k*(x-1)
x=(y+k)/k
y^2=4x=4*(y+k)/k
ky^2-4y-4k=0
yA+yB=4/k
yA*yB=-4
(yA-yB)^2=(yA+yB)^2-4yA*yB=(16+16k^2)/k^2
(xA-xB)^2+(yA-yB)^2=(1+1/k^2)*(16+16k^2)/k^2=AB^2=8^2
k=±1
y=±(x-1)