Question

13．在平面直角坐标系中，方程x²+y²=1所对应的图象经过伸缩变换$\left\{\begin{array}{l}x'=5x\\ y'=3y\end{array}\right.$后的图象所对应的方程为$\frac{x^2}{25}+\frac{y^2}{9}=1$．

Answer 1

分析 由伸缩变换$\left\{\begin{array}{l}x'=5x\\ y'=3y\end{array}\right.$，可得：$\left\{\begin{array}{l}{x=\frac{1}{5}x′}\\{y=\frac{1}{3}y′}\end{array}\right.$，代入方程x²+y²=1即可得出结论．

解答 解：由伸缩变换$\left\{\begin{array}{l}x'=5x\\ y'=3y\end{array}\right.$可得：$\left\{\begin{array}{l}{x=\frac{1}{5}x′}\\{y=\frac{1}{3}y′}\end{array}\right.$
代入方程x²+y²=1可得：$\frac{x{′}^{2}}{25}+\frac{y{′}^{2}}{9}$=1，即$\frac{x^2}{25}+\frac{y^2}{9}=1$．
故答案为：$\frac{x^2}{25}+\frac{y^2}{9}=1$．

点评 本题考查了坐标变换，考查了推理能力与计算能力，属于基础题．