(2011?河南模拟)已知抛物线C:y=mx2(m>0),焦点为F,直线2x-y+2=0交抛物线C于A、B两点,P是线段AB的

2025-02-26 15:05:37
推荐回答(1个)
回答1:

(1)∵抛物线C的焦点F(0,
1
4m
)

|RF|=yR+
1
4m
=2+
1
4m
=3
,得m=
1
4

(2)联立方程
y=mx2
2x?y+2=0

消去y得mx2-2x-2=0,设A(x1,mx12),B(x2,mx22),
x1+x2
2
m
x1?x2=?
2
m
(*),
∵P是线段AB的中点,∴P(
x1+x2
2
m
x
+m
x
2
)
,即P(
1
m
yp)
,∴Q(
1
m
1
m
)