已知曲线C的极坐标方程是ρ2（1+3sin2θ）=4，直线l的参数方程是
（1）求曲线C和直线l的直角坐标方程； （2）设点M为曲线C上任一点，求M到直线l的距离的最大值． |
（1）∵ρ2（1+3sin2θ）=4，
∴ρ2（cos2θ+4sin2θ）=4，
∴x2+4y2=4，
∴C：
+y2=1．x2 4
∵l的参数方程是
（t为参数），x=6-
t2 5 5 y=
t5 5
∴x=6-
•2 5 5
y，5
∴l：x+2y-6=0．
（2）设M（2cosθ，sinθ），
则M到直线l的距离d=
=|2cosθ+2sinθ-6| 5
，|2sin(θ+
)-6|π 4 5
∴当sin(θ+
)=-1，π 4
即θ=
，M(-5π 4
，-2
)时，2 2
dmax=
=6+2 2 5
．6
+25 10 5
| 完形填空。 | ||||
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