试题分析:把曲线的参数方程化为普通方程,由|AB|
2=|MA|•|MB|,可得|AB|等于圆的切线长,设出直线l的方程,求出弦心距d,再利用弦长公式求得|AB|,由此求得直线的斜率k的值,即可求得直线l的方程.
解:直线

的参数方程:

(

为参数),…………①
曲线

:

化为普通方程为

,…………②
将①代入②整理得:

,设

、

对应的参数分别为

,

,由

成等比数列得:

,

,

,

,
直线

的方程为:

点评:解决该试题的关键是把曲线的参数方程化为普通方程,由|AB|
2=|MA|•|MB|,可得|AB|等于圆的切线长,利用切割线定理得到,并结合勾股定理得到结论。