(1)直线l的方程可化为y=
| m |
| m2+1 |
| 4m |
| m2+1 |
| m |
| m2+1 |
即km2-m+k=0,∵△≥0,∴1-4k2≥0,
所以,斜率k的取值范围是[−
| 1 |
| 2 |
| 1 |
| 2 |
(2)不能.由(1知l的方程为y=k(x-4),其中|k|≤
| 1 |
| 2 |
圆C的圆心为C(4,-2),半径r=2;圆心C到直线l的距离d=
| 2 | ||
|
由|k|≤
| 1 |
| 2 |
| 4 | ||
|
| r |
| 2 |
从而,若l与圆C相交,则圆C截直线l所得的弦所对的圆心角小于
| 2π |
| 3 |
所以l不能将圆C分割成弧长的比值为
| 1 |
| 2 |
| 1 |
| 2 |
| m |
| m2+1 |
| 4m |
| m2+1 |
| m |
| m2+1 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 2 | ||
|
| 1 |
| 2 |
| 4 | ||
|
| r |
| 2 |
| 2π |
| 3 |
| 1 |
| 2 |