When the axes are rotated through an angle $\mathrm{\pi }$ /6, Find the transformed equation of ${\mathrm{x}}^2+2\sqrt{3}\mathrm{xy}-{\mathrm{y}}^2=2{\mathrm{a}}^2$.

Given original equation is ${\mathrm{x}}^2+2\sqrt{3}\mathrm{xy}-{\mathrm{y}}^2=2{\mathrm{a}}^2 ....\left(1\right)$
$\text{ Given }\mathrm{\theta }=\frac{\mathrm{\pi }}{6}={30}^0$
$\begin{array}{l}\mathrm{x}=\mathrm{Xcos}\mathrm{\theta }-\mathrm{Ysin}\mathrm{\theta }=\mathrm{Xcos}{30}^0-\mathrm{Ysin}{30}^0\\ =\mathrm{X}\frac{\sqrt{3}}{2}-\mathrm{Y}\frac{1}{2}=\frac{\sqrt{3}\mathrm{X}-\mathrm{Y}}{2}\\ \mathrm{y}=\mathrm{Xsin}\mathrm{\theta }+\mathrm{Ycos}\mathrm{\theta }=\mathrm{X}\left(\frac{1}{2}\right)+\mathrm{Y}\left(\frac{\sqrt{3}}{2}\right)=\frac{\mathrm{X}+\sqrt{3}\mathrm{Y}}{2}\end{array}$
Substitute x, y values in equation (1), we get the transformed equation
$\begin{array}{l}\Rightarrow {\left(\frac{\sqrt{3}\mathrm{X}-\mathrm{Y}}{2}\right)}^2+2\sqrt{3}\left(\frac{\sqrt{3}\mathrm{X}-\mathrm{Y}}{2}\right)\left(\frac{\mathrm{X}+\sqrt{3}\mathrm{Y}}{2}\right)-{\left(\frac{X+\sqrt{3}Y}{2}\right)}^2=2a^2\\ \Rightarrow \frac{3{\mathrm{X}}^2-2\sqrt{3}\mathrm{XY}+{\mathrm{Y}}^2}{4}+2\sqrt{3}\frac{\left(\sqrt{3}{\mathrm{X}}^2+3\mathrm{XY}-\mathrm{XY}-\sqrt{3}{Y}^2\right)}{4}-\frac{\left({\mathrm{X}}^2+3{\mathrm{Y}}^2+2\sqrt{3}\mathrm{XY}\right)}{4}=2{\mathrm{a}}^2\\ \\ \Rightarrow 3{\mathrm{X}}^2-2\sqrt{3}\mathrm{XY}+{\mathrm{Y}}^2+6{\mathrm{X}}^2+6\sqrt{3}\mathrm{XY}-2\sqrt{3}\mathrm{XY}-6{\mathrm{Y}}^2-{\mathrm{X}}^2-3{\mathrm{Y}}^2-2\sqrt{3}\mathrm{XY}=8{\mathrm{a}}^2\\ \\ \Rightarrow 8{\mathrm{X}}^2-8{\mathrm{Y}}^2=8{\mathrm{a}}^2\Rightarrow {\mathrm{X}}^2-{\mathrm{Y}}^2={\mathrm{a}}^2\end{array}$