问题 解答题
已知点F(1,0),直线l:x=2,设动点P到直线l的距离为d,已知|PF|=
2
2
d且
2
3
≤d≤
3
2

(1)求动点P的轨迹方程;
(2)若
PF
OF
=
1
3
,求向量
OP
OF
的夹角.
答案

(1)设动点P的坐标为(x,y),则 |PF|=

(x-1)2+y2
,d=|2-x|,

(x-1)2+y2
|2-x|
=
2
2

化简得

x2
2
+y2=1

2
3
≤d=2-x≤
3
2
1
2
≤x≤
4
3

即动点p的轨迹方程为

x2
2
+y2=1(
1
2
≤x≤
4
3
)

(2)∵

PF
=(1-x,-y),
OF
=(1,0),
OP
=(x,y)

PF
OF
=1-x=
1
3

x=

2
3
,代入
x2
2
+y2=1(
1
2
≤x≤
4
3
)
y=±
7
3

OP
=(
2
3
7
3
)或(
2
3
,-
7
3
)∴cos<
OP
OF
>=
OP
OF
|
OP
||
OF
|
=
2
11
11

OP
OF
的夹角为arccos
2
11
11

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