则CP⊥OQ,则
| CP |
| OQ |
∴(x-1,y)(x,y)=0,即(x−
| 1 |
| 2 |
| 1 |
| 4 |
(二)定义法:∵∠OPC=90°,动点P在以M(
| 1 |
| 2 |
∴所求点的轨迹方程为(x−
| 1 |
| 2 |
| 1 |
| 4 |
(三)参数法:设动弦PQ的方程为y=kx,由
|
得:(1+k2)x2-2x=0,设P(x1,y1),Q(x2,y2),
PQ的中点为(x,y),则:x=
| x1+x2 |
| 2 |
| 1 |
| 1+k2 |
| k |
| 1+k2 |
消去k得(x−
| 1 |
| 2 |
| 1 |
| 4 |
| CP |
| OQ |
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 4 |
|
| x1+x2 |
| 2 |
| 1 |
| 1+k2 |
| k |
| 1+k2 |
| 1 |
| 2 |
| 1 |
| 4 |