Question

17．欲将方程$\frac{x^2}{4}$+$\frac{y^2}{3}$=1所对应的图形变成方程x²+y²=1所对应的图形，需经过伸缩变换φ为（　　）

• A． $\left\{\begin{array}{l}x'=2x\\ y'=\sqrt{3}y\end{array}\right.$
• B． $\left\{\begin{array}{l}x'=\frac{1}{2}x\\ y'=\frac{{\sqrt{3}}}{3}y\end{array}\right.$
• C． $\left\{\begin{array}{l}x'=4x\\ y'=3y\end{array}\right.$
• D． $\left\{\begin{array}{l}{x′=\frac{1}{4}x}\\{y′=\frac{1}{3}y}\end{array}\right.$

Answer 1

分析 设伸缩变换φ为$\left\{\begin{array}{l}x'=hx\\ y'=ky\end{array}\right.，（h，k＞0）$，代入$\frac{x^2}{4}+\frac{y^2}{3}=1$，化简计算即可得到．

解答 解：设伸缩变换φ为$\left\{\begin{array}{l}x'=hx\\ y'=ky\end{array}\right.，（h，k＞0）$，
则$\left\{\begin{array}{l}x=\frac{x'}{h}\\ y=\frac{y'}{k}\end{array}\right.$，
代入$\frac{x^2}{4}+\frac{y^2}{3}=1$
得$\frac{x^2}{{4{h^2}}}+\frac{y^2}{{3{k^2}}}=1$，
∴$\left\{\begin{array}{l}4{h^2}=1\\ 3{k^2}=1\end{array}\right.⇒\left\{\begin{array}{l}h=\frac{1}{2}\\ k=\frac{{\sqrt{3}}}{3}\end{array}\right.$
故选：B

点评 本题考查了伸缩变换，关键是对变换公式的理解与运用，是基础题．