Suppose the axes X and Y are obtained by rotating the axes x and y an angle $\theta$. If the equation $x^2+2\sqrt{3}xy-y^2=4a^2$ is transformed to $X^2-Y^2=2a^2$ with respect to the XY axes, then $\theta$ is equal to

$$\begin{gathered} x=x\cos \theta -y\sin \theta \\ y=x\sin \theta +y\cos \theta \\ {(x\cos \theta -y\sin \theta )}^2+2\sqrt{3}(x\cos \theta -y\sin \theta ) \\ (x\sin \theta +y\cos \theta )-{(x\cos \theta +y\sin \theta )}^2-=4a^2 \end{gathered}$$
Is changing $x^2-y^2=2a^2$
$\begin{array}{l}\therefore -2xy\sin \theta \cos \theta +2\sqrt{3}(xy{\cos }^2\theta -\\ xy{\sin }^2\theta -2xy\sin \theta \cos \theta =0\end{array}$
$\begin{array}{l}-2\sin 2\theta +2\sqrt{3}\cos 2\theta =0\\ \Rightarrow \tan 2\theta =\sqrt{3}=\tan \frac{\pi }{3}\\ \theta =\frac{\pi }{6}\end{array}$