Q83 / 160 · TS EAMCET 2020 (14 Sep Shift 1) · Mathematics
Suppose the axes $X$ and $Y$ are obtained by rotating the axes $x$ and $y$ an angle $\theta$. If the equation $x^2+2 \sqrt{3} x y-y^2=4 a^2$ 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$90^{\circ}$
- Correct answerD$30^{\circ}$
Solution
$X$ and $Y$ are obtained by rotating $x$ and $y$ by angle
$\begin{array}{cc}
\because & x=X \cos \theta-Y \sin \theta \\
& y=X \sin \theta+Y \cos \theta \\
\because \quad & x^2+2 \sqrt{3} x y-y^2=4 a^2 \\
& (X \cos \theta-Y \sin \theta)^2+2 \sqrt{3}(X \cos \theta-Y \sin \theta) \\
& (X \sin \theta+Y \cos \theta)-(X \sin \theta+Y \cos \theta)^2=4 a^2
\end{array}$
is changing in $X^2-Y^2=2 a^2$
$\begin{gathered}
\therefore \quad-2 \times Y \sin \theta \cos \theta+2 \sqrt{3}\left(X Y \cos ^2 \theta-X Y \sin ^2 \theta\right) \\
-2 X Y \sin \theta \cos \theta=0 \\
-4 \sin \theta \cos \theta+2 \sqrt{3}\left(\cos ^2 \theta-\sin ^2 \theta\right)=0 \\
-2 \sin 2 \theta+2 \sqrt{3} \cos 2 \theta=0 \\
\tan \theta=\frac{1}{\sqrt{3}} \Rightarrow \theta=\frac{\pi}{6}=30^{\circ}
\end{gathered}$
Hence, the correct answer is option: 'D'.