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