题目内容
在直角坐标系xOy中,点P到两点(0,-
),(0,
)的距离之和等于4,设点P的轨迹为C,直线y=kx+1与C交于A,B两点.
(1)写出C的方程;
(2)若
⊥
,求k的值.
| 3 |
| 3 |
(1)写出C的方程;
(2)若
| OA |
| OB |
(1)设P(x,y),由椭圆定义可知,点P的轨迹C是以(0,-
),(0,
)为焦点,长半轴为2的椭圆,它的短半轴b=
=1,故曲线C的方程为x2+
=1.
(2)设A(x1,y1),B(x2,y2),其坐标满足
消去y并整理得
(k2+4)x2+2kx-3=0,
故x1+x2=-
,x1x2=-
.
∵
⊥
∴x1x2+y1y2=0.
∵y1y2=k2x1x2+k(x1+x2)+1,
∴x1x2+y1y2=-
-
-
+1=0,化简得-4k2+1=0,所以k=±
.
| 3 |
| 3 |
22-(
|
| y2 |
| 4 |
(2)设A(x1,y1),B(x2,y2),其坐标满足
|
(k2+4)x2+2kx-3=0,
故x1+x2=-
| 2k |
| k2+4 |
| 3 |
| k2+4 |
∵
| OA |
| OB |
∴x1x2+y1y2=0.
∵y1y2=k2x1x2+k(x1+x2)+1,
∴x1x2+y1y2=-
| 3 |
| k2+4 |
| 3k2 |
| k2+4 |
| 2k2 |
| k2+4 |
| 1 |
| 2 |
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