设O为坐标原点,抛物线y 2 =4x与过焦点的直线交于A、B两点,则 OA ? OB =

2025-06-25 06:53:31
推荐回答(1个)
回答1:

由题意知,抛物线y 2 =4x的焦点坐标为(1,0),
∴设直线AB的方程为y=k(x-1),
y 2 =4x
y=k(x-1)
?k 2 x 2 -(2k 2 +4)x+k 2 =0.
设出A(x 1 ,y 1 )、B(x 2 ,y 2
x 1 + x 2 =
2 k 2 +4
k 2
,x 1 x 2 =1.
∴y 1 ?y 2 =k(x 1 -1)?k(x 2 -1)=k 2 [x 1 x 2 -(x 1 +x 2 )+1].
OA
?
OB
=x 1 x 2 +y 1 y 2 =1+k 2 [2-
2 k 2 +4
k 2
]=-3.
当斜率不存在时仍然成立.
故选C.