Suppose the new axes X,Y are generated by rotating the coordinate axes x,y about the origin through an angle of ${30}^0$ in the anticlockwise
direction. Then the transformed equation of $x^2+ 2\sqrt{ 3}xy- y^2= 2a^2$ with respect to new axes X,Y is

$$\begin{gathered} \mathrm{Given} \mathrm{\theta }={30}^0 \\ \mathrm{Now} \mathrm{x}=X\cos {30}^0-Y \sin {30}^0 \\ =\frac{\sqrt{3}X}{2}-\frac{Y}{2}=\frac{\sqrt{3}X-Y}{2} \\ \mathrm{and} \mathrm{y}=\mathrm{Xsin}{30}^0+Y\cos {30}^0 \\ =\frac{X}{2}+\frac{\sqrt{3}Y}{2} \\ =\frac{X+\sqrt{3}Y}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{Now} \mathrm{transformed} \mathrm{equation} \mathrm{of} {\mathrm{x}}^2+2\sqrt{3}\mathrm{xy}-{\mathrm{y}}^2=2{\mathrm{a}}^2 \\ {\left(\frac{\sqrt{3}X-Y}{2}\right)}^2+2\sqrt{3}\left(\frac{\sqrt{3}X-Y}{2}\right)\left(\frac{X+Y}{2}\right)-{\left(\frac{X+\sqrt{3}Y}{2}\right)}^2=2{\mathrm{a}}^2 \\ \Rightarrow {\frac{(\sqrt{3}X-Y)}{4}}^2+2\sqrt{3}\frac{\left[\sqrt{3}{X}^2+2XY-\sqrt{3}{Y}^2\right]}{4}-\frac{{(X+\sqrt{3}Y)}^2}{4}=2{\mathrm{a}}^2 \\ \Rightarrow 3X^2+Y^2-2\sqrt{3}XY+6X^2+4\sqrt{3}XY-6Y^2-X^2-3Y^2-2\sqrt{3}XY=8{\mathrm{a}}^2 \\ \Rightarrow 8X^2-8Y^2 =8{\mathrm{a}}^2 \\ \Rightarrow X^2-Y^2={\mathrm{a}}^2 \end{gathered}$$