When  the axes are  rotated through  an angle of $\frac{\pi }{3}$ the transformed form of $7x^2+2 \sqrt{3} xy+9y^2–8 =0$ is

$$\begin{gathered} \mathrm{Given} \mathrm{\theta }=\frac{\mathrm{\pi }}{3}={60}^0 \\ \mathrm{let} (X,Y) \mathrm{be} \mathrm{new} \mathrm{coordinates} \mathrm{then} \\ \mathrm{x}=X \cos {60}^{°}-Y \sin {60}^{°} \\ =X\left(\frac{1}{2}\right)-Y\left(\frac{\sqrt{3}}{2}\right) \\ =\frac{X-\sqrt{3}Y}{2} \\ \mathrm{and} \mathrm{y}=X \sin {60}^0+Y\cos {60}^0 \\ =X\left(\frac{\sqrt{3}}{2}\right)+Y\left(\frac{1}{2}\right) \\ \mathrm{y}=\frac{\sqrt{3}X+Y}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{Given} \mathrm{original} \mathrm{equation} is 7{\mathrm{x}}^2+2\sqrt{3}\mathrm{xy}+9{\mathrm{y}}^2-8=0 \\ \Rightarrow 7{\left(\frac{X-\sqrt{3}Y}{2}\right)}^2+2\sqrt{3}\left(\frac{X-\sqrt{3}Y}{2}\right)\left(\frac{\sqrt{3}X+Y}{2}\right)+9{\left(\frac{\sqrt{3}X+Y}{2}\right)}^2-8=0 \\ \Rightarrow 7\left(\frac{X^2+3Y^2-2\sqrt{3}XY}{4}\right)+2\sqrt{3}\left[\frac{\sqrt{3}{X}^2+XY-3XY-\sqrt{3}{Y}^2}{4}\right]+9\left[\frac{3X^2+Y^2+2\sqrt{3}XY}{4}\right]-8=0 \\ \Rightarrow 7X^2+21Y^2-14\sqrt{3}XY+6X^2-4\sqrt{3}XY-6Y^2+27X^2+9Y^2+18\sqrt{3}XY-32=0 \\ \Rightarrow 40X^2+24Y^2-32=0 \\ \Rightarrow 8(5X^2+3Y^2-4)=0 \\ \Rightarrow 5X^2+3Y^2=4 \end{gathered}$$