Question

13．将圆x²+y²=1经过伸缩变换$\left\{\begin{array}{l}{x′=4x}\\{y′=3y}\end{array}\right.$后的曲线方程为$\frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}=1$．

Answer 1

分析 将$\left\{\begin{array}{l}{x′=4x}\\{y′=3y}\end{array}\right.$，整理可知：$\left\{\begin{array}{l}{x=\frac{x′}{4}}\\{y=\frac{y′}{3}}\end{array}\right.$，代入圆方程即可求得曲线方程．

解答 解：由$\left\{\begin{array}{l}{x′=4x}\\{y′=3y}\end{array}\right.$，整理可知：$\left\{\begin{array}{l}{x=\frac{x′}{4}}\\{y=\frac{y′}{3}}\end{array}\right.$，代入圆方程：$\frac{{x′}^{2}}{16}+\frac{{y′}^{2}}{9}=1$，
∴曲线方程：$\frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}=1$，
故答案为：$\frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}=1$．

点评 本题考查曲线方程的变换，考查转化思想，属于基础题．