Question

8．將圓的標(biāo)準(zhǔn)方程（x-1）²（y+3）²=4化成參數(shù)方程為$\left\{\begin{array}{l}{x=2cosα+1}\\{y=2sinα-3}\end{array}\right.$（α為參數(shù)）．

Answer 1

分析 根據(jù)同角三角函數(shù)的關(guān)系設(shè)參數(shù)解出x，y得出答案．

解答 解：∵（x-1）²（y+3）²=4，∴$\frac{（x-1）^{2}}{4}+\frac{（y+3）^{2}}{4}=1$．
令$\frac{x-1}{2}=cosα$，$\frac{y+3}{2}=sinα$，則x=2cosα+1，y=2sinα-3．
∴圓的參數(shù)方程為$\left\{\begin{array}{l}{x=2cosα+1}\\{y=2sinα-3}\end{array}\right.$（α為參數(shù)）．
故答案為：$\left\{\begin{array}{l}{x=2cosα+1}\\{y=2sinα-3}\end{array}\right.$（α為參數(shù)）．

點(diǎn)評(píng) 本題考查了圓的參數(shù)方程，屬于基礎(chǔ)題．