Question

17．欲將方程$\frac{x^2}{4}$+$\frac{y^2}{3}$=1所對(duì)應(yīng)的圖形變成方程x²+y²=1所對(duì)應(yīng)的圖形，需經(jīng)過(guò)伸縮變換φ為（　　）

• 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 設(shè)伸縮變換φ為$\left\{\begin{array}{l}x'=hx\\ y'=ky\end{array}\right.，（h，k＞0）$，代入$\frac{x^2}{4}+\frac{y^2}{3}=1$，化簡(jiǎn)計(jì)算即可得到．

解答 解：設(shè)伸縮變換φ為$\left\{\begin{array}{l}x'=hx\\ y'=ky\end{array}\right.，（h，k＞0）$，
則$\left\{\begin{array}{l}x=\frac{x'}{h}\\ y=\frac{y'}{k}\end{array}\right.$，
代入$\frac{x^2}{4}+\frac{y^2}{3}=1$
得$\frac{x^2}{{4{h^2}}}+\frac{y^2}{{3{k^2}}}=1$，
∴$\left\{\begin{array}{l}4{h^2}=1\\ 3{k^2}=1\end{array}\right.⇒\left\{\begin{array}{l}h=\frac{1}{2}\\ k=\frac{{\sqrt{3}}}{3}\end{array}\right.$
故選：B

點(diǎn)評(píng) 本題考查了伸縮變換，關(guān)鍵是對(duì)變換公式的理解與運(yùn)用，是基礎(chǔ)題．