已知抛物线y^2=4x的焦点为F,过点F的直线l与抛物线交于A.B两点|AB|=8 求AB的直线方程
问题描述：
已知抛物线y^2=4x的焦点为F,过点F的直线l与抛物线交于A.B两点|AB|=8 求AB的直线方程
答
F(1,0)
AB:y=k(x-1)
x=(k+y)/k
y^2=4x=4*(k+y)/k
ky^2-4y-4k=0
yA+yB=4/k
yA*yB=-4
(yA-yB)^2=(yA+yB)^2-4yA*yB=16/k^2+16
(xA-xB)^2+(yA-yB)^2=(1+1/k^2)*(16/k^2+16)=AB^2=8^2
k=±1
AB:y=±(x-1)