The equation,


represents a pair of intersecting lines, so
$$
\begin{aligned}
& \text { (3) }(2)(2)+7(g)(f)-3 f^2-2 g^2-2\left(\frac{7}{2}\right)^2=0 \\
& \Rightarrow \quad 24+14 g f-6 f^2-4 g^2-49=0
\end{aligned}
$$

To get the point of intersection of pair of straight lines, we have to differentiate partially the Eq. (i) with respect to $x$ and $y$ respectively.
then we are getting

Now, point of intersection of eqs. (iii) and (iv) is the point of intersection of pair of straight line Eq. (i).
$$
\left(\frac{14 f-8 g}{25}, \frac{14 g-12 f}{25}\right)
$$
Now, according to the question,
$$
\begin{aligned}
& \left(\frac{14 f-8 g}{25}\right)^2+\left(\frac{14 g-12 f}{25}\right)^2=\frac{2}{5} \\
\Rightarrow \quad 196 f^2+64 g^2-224 f g+196 g^2 & +144 f^2 \\
\Rightarrow \quad & \quad 260 g^2+340 f^2-560 g f=250
\end{aligned}
$$

From Eqs. (ii) and (v)
$$
\begin{gathered}
26 g^2+34 f^2-56 g f=25 \\
16 g^2+24 f^2-56 g f=-100 \\
-\frac{-\quad+\quad+}{10 g^2+10 f^2=125}+ \\
g^2+f^2=\frac{25}{2}
\end{gathered}
$$