Question

9．参数方程$\left\{{\begin{array}{l}{x=3+sinθ}\\{y=cosθ}\end{array}}\right.$（θ为参数）化为普通方程是（x-3）²+y²=1．

Answer 1

分析 由参数方程$\left\{{\begin{array}{l}{x=3+sinθ}\\{y=cosθ}\end{array}}\right.$可得$\left\{\begin{array}{l}{sinθ=x-3}\\{cosθ=y}\end{array}\right.$，结合sin²θ+cos²θ=1可得答案．

解答 解：由参数方程$\left\{{\begin{array}{l}{x=3+sinθ}\\{y=cosθ}\end{array}}\right.$可得$\left\{\begin{array}{l}{sinθ=x-3}\\{cosθ=y}\end{array}\right.$，
两边平方作和得（x-3）²+y²=1．
故答案为：（x-3）²+y²=1．

点评 本题主要考查参数方程与普通方程的相互转化，属于基础题．