If the coordinate axes are rotated through an angle $\frac{\pi}{6}$ about the origin, then the transformed…
If the coordinate axes are rotated through an angle $\frac{\pi}{6}$ about the origin, then the transformed equation of $\sqrt{3} x^2-4 x y+\sqrt{3} y^2=0$ is
$\sqrt{3} y^2+x y=0$
$x^2 - y^2 = 0$
$\sqrt{3} y^2-x y=0$
$\sqrt{3} y^2- 2x y=0$
Solution
Given equation, $\sqrt{3} x^2-4 x y+\sqrt{3} y^2=0$
Coordinate axes are rotated through an angle of $\frac{\pi}{6}$ about the origin
Now, axis is $x \cos \frac{\pi}{6} \quad y \cos \frac{\pi}{6}$
and $x \cos \frac{\pi}{6}+y \cos \frac{\pi}{6}$
Thus, equation becomes $\sqrt{3}\left(x \cos \frac{\pi}{6}-y \cos \frac{\pi}{6}\right)^2$
$\begin{aligned}
& -4\left(x \cos \frac{\pi}{6}-y \cos \frac{\pi}{6}\right) \\
& \left(x \cos \frac{\pi}{6}+y \cos \frac{\pi}{6}\right)+\sqrt{3}\left(x \cos \frac{\pi}{6}+y \cos \frac{\pi}{6}\right)^2=0 \\
& \Rightarrow \sqrt{3}\left(\frac{\sqrt{3} x}{2}-\frac{1}{2} y\right)^2-4\left(\frac{\sqrt{3} x-y}{2}\right) \\
& \left(\frac{x+\sqrt{3} y}{2}\right)+\sqrt{3}\left(\frac{x+\sqrt{3} y}{2}\right)^2=0 \\
& \Rightarrow \sqrt{3}\left(3 x^2-2 \sqrt{3} x y+y^2\right)-4\left(\sqrt{3} x^2+3 x y-x y-\sqrt{3} y^2\right) \\
& \Rightarrow \quad 3 \sqrt{3} x^2-6 x y+\sqrt{3} y^2-4 \sqrt{3} x^2-12 x y \\
& \quad+4 x y+4 \sqrt{3} y^2+\sqrt{3} x^2+6 x y+3 \sqrt{3} y^2=0 \\
& \Rightarrow \quad 8 \sqrt{3} y^2-8 x y=0 \\
& \therefore \quad \sqrt{3} y^2-x y=0
\end{aligned}$