(1)∵抛物线C的焦点F(0,
),1 4m
∴|RF|=yR+
=2+1 4m
=3,得m=1 4m
.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),
则
(*),
x1+x2=
2 m
x1?x2=?
2 m
∵P是线段AB的中点,∴P(
,
x1+x2
2
),即P(m
+m
x
x
2
,yp),∴Q(1 m
,1 m
),1 m
得