一道高二抛物线题

抛物线y^2＝4x的弦AB＝4√3
焦点F(1,0)
设AB:x=ky+b
y^2=4x=4(ky+b)
y^2-4ky-4b=0
yA+yB=4k
yA*yB=-4b
yA^2=4xA......(1)
yB^2=4xB......(2)
(1)-(2):
yA^2-yB^2=4*(xA-xB)
(yA+yB)*(yA-yB)=4(xA-xB)
(yA+yB)*(yA-yB)/(xA-xB)=4
(yA+yB)*k=4
yA+yB=4/k=4k
k=±1
k=1或k=-1，结果相同，取k=1
(yA-yB)^2=(yA+yB)^2-4yA*yB=16k^2+16b=16+16b
(xA-xB)^2=(yA-yB)^2
AB^2=2(yA-yB)^2
(4√3)^2=2(16+16b)
48=2(16+16b)
b=0.5
AB:x=y+0.5
焦点F(1,0)到AB的距离=|1-0.5|/√2=√2/4
经检验正确，所以不是√2

点差法