Question

Suppose the axes $X$ and $Y$ are obtained by rotating the axes $x$ and $y$ by 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

• A. $45^{\circ }$
• B. $60^{\circ }$
• C. $90^{\circ }$
• D. $30^{\circ }$

Answer 1

Answer: (D)

$X$ and $Y$ are obtained by rotating $x$ and $y$ by angle
$\because x=X\cos \theta -Y\sin \theta$
$y=X\sin \theta +Y\cos \theta$
$\because x^2+2\sqrt{3}xy-y^2=4a^2$
$(X\cos \theta -Y\sin \theta {)}^2+2\sqrt{3}(X\cos \theta -Y\sin \theta )$
$(X\sin \theta +Y\cos \theta )-(X\sin \theta +Y\cos \theta {)}^2=4a^2$
is changing in $X^2-Y^2=2a^2$
$\therefore -2\times Y\sin \theta \cos \theta +2\sqrt{3}\left(XY{\cos }^2\theta -XY{\sin }^2\theta \right)$
$-2XY\sin \theta \cos \theta =0$
$-4\sin \theta \cos \theta +2\sqrt{3}\left({\cos }^2\theta -{\sin }^2\theta \right)=0$
$-2\sin 2\theta +2\sqrt{3}\cos 2\theta =0$
$\tan \theta =\frac{1}{\sqrt{3}}$
$\Rightarrow \theta =\frac{\pi }{6}=30^{\circ }$