想必現(xiàn)在有很多小伙伴對于在直角坐標(biāo)系$xOy$中，把矩形$B=\left ( {\stackrel {\dfrac {1} {2}} {0}\stackrel {0} {1}} \right )$確定的壓縮變換$σ$與矩陣$A=\left ( {\stackrel {0} {1}\stackrel {-1} {0}} \right )$確定的旋轉(zhuǎn)變換${R}_{9{0}^{\circ }}$進行復(fù)合，得到復(fù)合變換${R}_{9{0}^{\circ }}\cdot σ$.$\left ( {1} \right )$求復(fù)合變換${R}_{9{0}^{\circ }}\cdot σ$的坐標(biāo)變換公式；$\left ( {2} \right )$求圓$C：{x}^{2}+{y}^{2}=1$在復(fù)合變換${R}_{9{0}^{\circ }}\cdot σ$的作用下所得曲線${C}^{'}$的方程.","title_text":"在直角坐標(biāo)系$xOy$中，把矩形$B=\left ( {\stackrel {\dfrac {1} {2}} {0}\stackrel {0} {1}} \right )$確定的壓縮變換$σ$與矩陣$A=\left ( {\stackrel {0} {1}\stackrel {-1} {0}} \right )$確定的旋轉(zhuǎn)變換${R}_{9{0}^{\circ }}$進行復(fù)合，得到復(fù)合變換${R}_{9{0}^{\circ }}\cdot σ$.$\left ( {1} \right )$求復(fù)合變換${R}_{9{0}^{\circ }}\cdot σ$的坐標(biāo)變換公式；$\left ( {2} \right )$求圓$C：{x}^{2}+{y}^{2}=1$在復(fù)合變換${R}_{9{0}^{\circ }}\cdot σ$的作用下所得曲線${C}^{'}$的方程.方面的知識都比較想要了解，那么今天小好小編就為大家收集了一些關(guān)于在直角坐標(biāo)系$xOy$中，把矩形$B=\left ( {\stackrel {\dfrac {1} {2}} {0}\stackrel {0} {1}} \right )$確定的壓縮變換$σ$與矩陣$A=\left ( {\stackrel {0} {1}\stackrel {-1} {0}} \right )$確定的旋轉(zhuǎn)變換${R}_{9{0}^{\circ }}$進行復(fù)合，得到復(fù)合變換${R}_{9{0}^{\circ }}\cdot σ$.$\left ( {1} \right )$求復(fù)合變換${R}_{9{0}^{\circ }}\cdot σ$的坐標(biāo)變換公式；$\left ( {2} \right )$求圓$C：{x}^{2}+{y}^{2}=1$在復(fù)合變換${R}_{9{0}^{\circ }}\cdot σ$的作用下所得曲線${C}^{'}$的方程.","title_text":"在直角坐標(biāo)系$xOy$中，把矩形$B=\left ( {\stackrel {\dfrac {1} {2}} {0}\stackrel {0} {1}} \right )$確定的壓縮變換$σ$與矩陣$A=\left ( {\stackrel {0} {1}\stackrel {-1} {0}} \right )$確定的旋轉(zhuǎn)變換${R}_{9{0}^{\circ }}$進行復(fù)合，得到復(fù)合變換${R}_{9{0}^{\circ }}\cdot σ$.$\left ( {1} \right )$求復(fù)合變換${R}_{9{0}^{\circ }}\cdot σ$的坐標(biāo)變換公式；$\left ( {2} \right )$求圓$C：{x}^{2}+{y}^{2}=1$在復(fù)合變換${R}_{9{0}^{\circ }}\cdot σ$的作用下所得曲線${C}^{'}$的方程.方面的知識分享給大家，希望大家會喜歡哦。

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 $復(fù)合變換${R}_{9{0}^{circ }}cdot σ$對應(yīng)的矩陣為$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 $復(fù)合變換${R}_{9{0}^{circ }}cdot σ$的坐標(biāo)變換公式為$left { {{begin{array}{ll} {{x}^{'}=-y} {{y}^{'}=dfrac {1} {2}x} end{array}}} right .$.

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

5、$left ( {2} right )$設(shè)圓$C$上任意一點$Pleft ( {x，y} right )$在變換${R}_{9{0}^{circ }}cdot σ$的作用下所得的點為${P}^{'}=left ( {{x}^{'}，{y}^{'}} right )$，

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

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

9、綜上所述：答案為${x}^{2}+4{y}^{2}=1$.

本文到此結(jié)束，希望對大家有所幫助。

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