$\therefore$ We are given that axis rotated through an angle of $30^{\circ}$. Now let new co-ordinates of the curve is $(\mathrm{X}, \mathrm{Y})$
Now $\mathrm{X}=x \cos \alpha-\mathrm{y} \sin \alpha, \mathrm{Y}=\mathrm{x} \sin \alpha+\mathrm{y} \cos \alpha$
$$
\begin{aligned}
& \Rightarrow \mathrm{X}=x \cos \left(\frac{\pi}{6}\right)-\mathrm{y} \sin \left(\frac{\pi}{6}\right) \Rightarrow \mathrm{X}=\frac{\sqrt{3}}{2} \mathrm{x}-\frac{1}{2} \mathrm{y} \\
& \mathrm{Y}=x \sin \left(\frac{\pi}{6}\right)+\mathrm{y} \cos \left(\frac{\pi}{6}\right) \Rightarrow \mathrm{y}=\frac{1}{2} \mathrm{x}+\frac{\sqrt{3}}{2} \mathrm{y}
\end{aligned}
$$
$\therefore$ New equation of the curve is:-
$$
\begin{aligned}
4 X^2+ & 12 X Y+9 Y^2+6 X+9 Y+2=0 \\
I & \Rightarrow 4\left(\frac{\sqrt{3}}{2} x-\frac{1}{2} y\right)^2+12\left(\frac{\sqrt{3}}{2} x-\frac{1}{2} y\right)\left(\frac{1}{2} x+\frac{\sqrt{3}}{2} y\right) \\
& +9\left(\frac{1}{2} x+\frac{\sqrt{3}}{2} y\right)^2+6\left(\frac{\sqrt{3}}{2} x-\frac{1}{2} y\right)+9\left(\frac{1}{2} x+\frac{\sqrt{3}}{2} y\right)+2=0
\end{aligned}
$$
Now $2 g=\frac{6 \sqrt{3}}{2}+\frac{9}{2} \Rightarrow 2 g=\frac{6 \sqrt{3}+9}{2}$
$$
2 \mathrm{f}=\frac{-6}{2}+\frac{9 \sqrt{3}}{2} \Rightarrow 2 \mathrm{f}=\frac{-6+9 \sqrt{3}}{2}
$$
Now, $\frac{2 \mathrm{~g}}{2 \mathrm{f}}=\frac{6 \sqrt{3}+9}{2} \times \frac{2}{-6+9 \sqrt{3}}$
$$
\Rightarrow \quad \frac{\mathrm{g}}{\mathrm{f}}=\frac{3(2 \sqrt{3}+3)}{3(3 \sqrt{3}-2)} \Rightarrow \frac{\mathrm{g}}{\mathrm{f}}=\frac{2 \sqrt{3}+3}{3 \sqrt{3}-2}
$$