想必现在有很多小伙伴对于在直角坐标系$xOy$中，把矩形$B=\left ( {\stackrel {\dfrac {1} {2}} {0}\stackrel {0} {1}} \right )$确定的压缩变换$σ$与矩阵$A=\left ( {\stackrel {0} {1}\stackrel {-1} {0}} \right )$确定的旋转变换${R}_{9{0}^{\circ }}$进行复合，得到复合变换${R}_{9{0}^{\circ }}\cdot σ$.$\left ( {1} \right )$求复合变换${R}_{9{0}^{\circ }}\cdot σ$的坐标变换公式；$\left ( {2} \right )$求圆$C：{x}^{2}+{y}^{2}=1$在复合变换${R}_{9{0}^{\circ }}\cdot σ$的作用下所得曲线${C}^{'}$的方程.","title_text":"在直角坐标系$xOy$中，把矩形$B=\left ( {\stackrel {\dfrac {1} {2}} {0}\stackrel {0} {1}} \right )$确定的压缩变换$σ$与矩阵$A=\left ( {\stackrel {0} {1}\stackrel {-1} {0}} \right )$确定的旋转变换${R}_{9{0}^{\circ }}$进行复合，得到复合变换${R}_{9{0}^{\circ }}\cdot σ$.$\left ( {1} \right )$求复合变换${R}_{9{0}^{\circ }}\cdot σ$的坐标变换公式；$\left ( {2} \right )$求圆$C：{x}^{2}+{y}^{2}=1$在复合变换${R}_{9{0}^{\circ }}\cdot σ$的作用下所得曲线${C}^{'}$的方程.方面的知识都比较想要了解，那么今天小好小编就为大家收集了一些关于在直角坐标系$xOy$中，把矩形$B=\left ( {\stackrel {\dfrac {1} {2}} {0}\stackrel {0} {1}} \right )$确定的压缩变换$σ$与矩阵$A=\left ( {\stackrel {0} {1}\stackrel {-1} {0}} \right )$确定的旋转变换${R}_{9{0}^{\circ }}$进行复合，得到复合变换${R}_{9{0}^{\circ }}\cdot σ$.$\left ( {1} \right )$求复合变换${R}_{9{0}^{\circ }}\cdot σ$的坐标变换公式；$\left ( {2} \right )$求圆$C：{x}^{2}+{y}^{2}=1$在复合变换${R}_{9{0}^{\circ }}\cdot σ$的作用下所得曲线${C}^{'}$的方程.","title_text":"在直角坐标系$xOy$中，把矩形$B=\left ( {\stackrel {\dfrac {1} {2}} {0}\stackrel {0} {1}} \right )$确定的压缩变换$σ$与矩阵$A=\left ( {\stackrel {0} {1}\stackrel {-1} {0}} \right )$确定的旋转变换${R}_{9{0}^{\circ }}$进行复合，得到复合变换${R}_{9{0}^{\circ }}\cdot σ$.$\left ( {1} \right )$求复合变换${R}_{9{0}^{\circ }}\cdot σ$的坐标变换公式；$\left ( {2} \right )$求圆$C：{x}^{2}+{y}^{2}=1$在复合变换${R}_{9{0}^{\circ }}\cdot σ$的作用下所得曲线${C}^{'}$的方程.方面的知识分享给大家，希望大家会喜欢哦。

1、$left ( {1} right )because A=left ( {stackrel {0} {1}stackrel {-1} {0}} right )$，$B=left ( {stackrel {dfrac {1} {2}} {0}stackrel {0} {1}} right )$.

2、$therefore $复合变换${R}_{9{0}^{circ }}cdot σ$对应的矩阵为$AB=left ( {stackrel {0} {1}stackrel {-1} {0}} right )left ( {stackrel {dfrac {1} {2}} {0}stackrel {0} {1}} right )=left ( {stackrel {0} {dfrac {1} {2}}stackrel {-1} {0}} right )$.

3、$therefore $复合变换${R}_{9{0}^{circ }}cdot σ$的坐标变换公式为$left { {{begin{array}{ll} {{x}^{'}=-y} {{y}^{'}=dfrac {1} {2}x} end{array}}} right .$.

4、综上所述：答案为$left { {{begin{array}{ll} {{x}^{'}=-y} {{y}^{'}=dfrac {1} {2}x} end{array}}} right .$.

5、$left ( {2} right )$设圆$C$上任意一点$Pleft ( {x，y} right )$在变换${R}_{9{0}^{circ }}cdot σ$的作用下所得的点为${P}^{'}=left ( {{x}^{'}，{y}^{'}} right )$，

6、由$left ( {1} right )$得$left { {{begin{array}{ll} {{x}^{'}=-y} {{y}^{'}=dfrac {1} {2}x} end{array}}} right .$，即$left { {{begin{array}{ll} {x=2{x}^{'}} {y=-{x}^{'}} end{array}}} right .left ( {ast } right )$.

7、将$left ( {ast } right )$代入圆$C：{x}^{2}+{y}^{2}=1$，得$left ( {{2y}^{'}} right )^{2}+left ( {{-x}^{'}} right )^{2}=1$，

8、$therefore $曲线${C}^{'}$的方程是${x}^{2}+4{y}^{2}=1$.

9、综上所述：答案为${x}^{2}+4{y}^{2}=1$.

本文到此结束，希望对大家有所帮助。

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