Question Number 187335 by Humble last updated on 16/Feb/23
![Apply the rotation of coordinates given by the following matrix to the equation xy=1; what is the equation in th uv coordinate system? [(u),(v) ]= [((cos45 sin45)),((−sin45 cos45)) ] [(x),(y) ]](https://www.tinkutara.com/question/Q187335.png)
$${Apply}\:{the}\:{rotation}\:{of}\:{coordinates}\:{given} \\ $$$${by}\:{the}\:{following}\:{matrix}\:{to}\:{the}\:{equation}\: \\ $$$${xy}=\mathrm{1};\:{what}\:{is}\:{the}\:{equation}\:{in}\:{th}\:{uv}\:{coordinate}\: \\ $$$${system}? \\ $$$$\begin{bmatrix}{{u}}\\{{v}}\end{bmatrix}=\begin{bmatrix}{{cos}\mathrm{45}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{sin}\mathrm{45}}\\{−{sin}\mathrm{45}\:\:\:\:\:\:\:\:\:\:{cos}\mathrm{45}}\end{bmatrix}\begin{bmatrix}{{x}}\\{{y}}\end{bmatrix} \\ $$
Answered by anurup last updated on 16/Feb/23
![[(x),(y) ]= [((1/( (√2))),(1/( (√2)))),((−(1/( (√2)))),(1/( (√2)))) ]^(−1) [(u),(v) ] = [((1/( (√2))),(−(1/( (√2))))),((1/( (√2))),(1/( (√2)))) ] [(u),(v) ] ∴x=(1/( (√2) ))(u−v) y=(1/( (√2)))(u+v) xy=1⇒(1/2)(u^2 −v^2 )=1⇒u^2 −v^2 =2](https://www.tinkutara.com/question/Q187345.png)
$$\begin{bmatrix}{{x}}\\{{y}}\end{bmatrix}=\:\begin{bmatrix}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}&{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\\{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}&{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\end{bmatrix}^{−\mathrm{1}} \begin{bmatrix}{{u}}\\{{v}}\end{bmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\begin{bmatrix}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}&{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\\{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}&{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\end{bmatrix}\begin{bmatrix}{{u}}\\{{v}}\end{bmatrix} \\ $$$$\therefore{x}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\left({u}−{v}\right) \\ $$$${y}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left({u}+{v}\right) \\ $$$${xy}=\mathrm{1}\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\left({u}^{\mathrm{2}} −{v}^{\mathrm{2}} \right)=\mathrm{1}\Rightarrow{u}^{\mathrm{2}} −{v}^{\mathrm{2}} =\mathrm{2} \\ $$$$ \\ $$
Commented by Humble last updated on 16/Feb/23

$${Thank}\:{you},\:{sir} \\ $$