设F为抛物线y2=4x的焦点,A、B为该抛物线上两点,若FA+2FB=0,则|FA|+2|FB|等于 _ .
问题描述:
设F为抛物线y2=4x的焦点,A、B为该抛物线上两点,若
+2
FA
=
FB
,则|0
|+2|
FA
|等于 ___ .
FB
答
过A,B两点分别作准线的垂线,再过B作AC的垂线,垂足为E,
设BF=m,则BD=m,
∵
+2FA
=FB
,0
∴AC=AF=2m,
如图,在直角三角形ABE中,
AE=AC-BD=2m-m=m,
AB=3m,∴cos∠BAE=
=AE AB
1 3
∴直线AB的斜率为:k=tan∠BAE=2
2
∴直线AB的方程为:y=2
(x-1)
2
将其代入抛物线的方程化简得:2x2-5x+2=0
∴x1=2,x2=
1 2
∴A(2,2
).B(
2
,1 2
),又F(1,0)
2
则|
|+2|FA
|=2FB
=6.
1+8
故答案为:6.