If the coordinate axes are rotated through an angle $\frac{\pi }{6}$ about the origin, then the transformed equation of $\sqrt{3}{x}^2-4xy+\sqrt{3}{y}^2=0$  is

$$\begin{gathered} \mathrm{Given} \mathrm{\theta }=\frac{\mathrm{\pi }}{6}={30}^0 \\ \mathrm{Now} \mathrm{x}=\mathrm{Xcos\theta }-\mathrm{Ysin\theta } \\ \mathrm{x}=X\left(\frac{\sqrt{3}}{2}\right)-Y\left(\frac{1}{2}\right) \\ \mathrm{x}=\frac{\sqrt{3} X-Y}{2} \\ \mathrm{and} \mathrm{y}=X \mathrm{sin\theta }+Y\mathrm{cos\theta } \\ =X \sin {30}^0+Y\cos {30}^0 \\ =X\left(\frac{1}{2}\right) +Y\left(\frac{\sqrt{3}}{2}\right) \\ \mathrm{y}=\frac{X+\sqrt{3}Y}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{Now} \mathrm{given} \mathrm{original} \mathrm{equation} is \sqrt{3} {\mathrm{x}}^2-4\mathrm{xy}+\sqrt{3} {\mathrm{y}}^2=0 \\ It\text{'}s transformed equation is \\ \Rightarrow \sqrt{3}\frac{{(\sqrt{3}X-Y)}^2}{4}-4\frac{(\sqrt{3}X-Y)(X+\sqrt{3}Y)}{4}+\sqrt{3}\frac{{(X+\sqrt{3}Y)}^2}{4}=0 \\ \Rightarrow \sqrt{3}(3X^2+Y^2-2\sqrt{3}XY)-4(\sqrt{3}{X}^2+3XY-XY-\sqrt{3}{Y}^2)+\sqrt{3}(X^2+3Y^2+2\sqrt{3}XY)=0 \\ \Rightarrow 3\sqrt{3}{X}^2+\sqrt{3}{Y}^2-6XY-4\sqrt{3}{X}^2-8XY+4\sqrt{3}{Y}^2+\sqrt{3}{X}^2+3\sqrt{3}{Y}^2+6XY=0 \\ \Rightarrow 8\sqrt{3}{Y}^2-8XY=0 \\ \Rightarrow \sqrt{3}{Y}^2-XY=0 \end{gathered}$$