题目内容

15.在同一坐标系中,将椭圆$\frac{{x}^{2}}{16}$+$\frac{{y}^{2}}{25}$=1变换成单位圆的伸缩变换是(  )
A.φ:$\left\{\begin{array}{l}{x′=5x}\\{{y}^{′}=4y}\end{array}\right.$B.φ:$\left\{\begin{array}{l}{{x}^{′}=4x}\\{{y}^{′}=5y}\end{array}\right.$
C.φ:$\left\{\begin{array}{l}{{x}^{′}=\frac{1}{4}x}\\{{y}^{′}=\frac{1}{5}y}\end{array}\right.$D.φ:$\left\{\begin{array}{l}{{x}^{′}=\frac{1}{5}x}\\{{y}^{′}=\frac{1}{4}y}\end{array}\right.$

分析 设$\left\{\begin{array}{l}{x′=λx}\\{y′=μy}\end{array}\right.$,得$\left\{\begin{array}{l}{x=\frac{x′}{λ}}\\{y=\frac{y′}{μ}}\end{array}\right.$,代入$\frac{{x}^{2}}{16}$+$\frac{{y}^{2}}{25}$=1后求得λ,μ值得答案.

解答 解:设$\left\{\begin{array}{l}{x′=λx}\\{y′=μy}\end{array}\right.$,则$\left\{\begin{array}{l}{x=\frac{x′}{λ}}\\{y=\frac{y′}{μ}}\end{array}\right.$,代入$\frac{{x}^{2}}{16}$+$\frac{{y}^{2}}{25}$=1得:
$\frac{(x′)^{2}}{16{λ}^{2}}+\frac{(y′)^{2}}{25{μ}^{2}}=1$,
∵椭圆$\frac{{x}^{2}}{16}$+$\frac{{y}^{2}}{25}$=1变换成单位圆,
∴16λ2=25μ2=1,即$λ=\frac{1}{4},μ=\frac{1}{5}$.
则φ:$\left\{\begin{array}{l}{x′=\frac{1}{4}x}\\{y′=\frac{1}{5}y}\end{array}\right.$.
故选:C.

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.

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