题目内容

在平面直角坐标系中,O是原点,
OA
=(1、0),P是平面内的动点,如|
OP
-
OA
|=|
OP
OA
|
,则P点的轨迹是(  )
分析:设P(x,y),则
OP
=(x,y)
OP
 -
OA
=( x-1,y)
,由|
OP
-
OA
|=|
OP
OA
|
,知
(x-1)2+y2
=|x|
,由此能求出P点的轨迹.
解答:解:设P(x,y),则
OP
=(x,y)

OA
=(1,0)

OP
 -
OA
=( x-1,y)

|
OP
-
OA
|=|
OP
OA
|

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

两边平方,得x2-2x+1+y2=x2
整理,得y2=2x-1.
∴P点的轨迹是抛物线.
故选D.
点评:本题考查向量在几何中的应用,是基础题.解题时要认真审题,仔细解答,注意合理地进行等价转化.
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