Question

Suppose the axes $X$ and $Y$ are obtained by rotating the axes $x$ and $y$ by an angle $\theta$. If the equation $x^{2}+2 \sqrt{3} x y-y^{2}=4 a^{2}$ is transformed to $X ^{2}- Y ^{2}=2 a ^{2}$ with respect to the $X Y$ - 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
$\because x=X \cos \theta-Y \sin \theta$
$y=X \sin \theta+Y \cos \theta$
$\because x^{2}+2 \sqrt{3} x y-y^{2}=4 a^{2}$
$(X \cos \theta-Y \sin \theta)^{2}+2 \sqrt{3}(X \cos \theta-Y \sin \theta)$
$(X \sin \theta+Y \cos \theta)-(X \sin \theta+Y \cos \theta)^{2}=4 a^{2}$
is changing in $X^{2}-Y^{2}=2 a^{2}$
$\therefore -2 \times Y \sin \theta \cos \theta+2 \sqrt{3}\left(X Y \cos ^{2} \theta-X Y \sin ^{2} \theta\right)$
$-2 X Y \sin \theta \cos \theta=0$
$-4 \sin \theta \cos \theta+2 \sqrt{3}\left(\cos ^{2} \theta-\sin ^{2} \theta\right)=0$
$-2 \sin 2 \theta+2 \sqrt{3} \cos 2 \theta=0$
$\tan \theta=\frac{1}{\sqrt{3}} $
$\Rightarrow \theta=\frac{\pi}{6}=30^{\circ}$