and equation of the curve is

Making Eq. (ii) homogeneous by Eq. (i), we get the equation of the lines joining the origin to the point of intersection of Eqs. (i) and (ii).
$$
\begin{aligned}
& x^2+y^2+2 h x y+(g x+f y)(x+y)+1(x+y)^2=0 \\
& \begin{aligned}
\Rightarrow x^2+y^2+2 h x y+g x^2+g x y \\
\quad+f x y+f y^2+x^2+y^2+2 x y=0
\end{aligned} \\
& \Rightarrow(2+g) x^2+(2+f) y^2+x y(g+f+2 h+2)=0 \quad \ldots \text { (iii) }
\end{aligned}
$$
Lines denoted by Eq. (iii) will be perpendicular to each other if
Coefficient of $x^2+$ Coefficient of $y^2=0$
i.e.
$$
(2+g)+(2+f)=0 \Rightarrow g+f+4=0
$$
Locus of $(g, f)$ is $x+y+4=0$
i.e. $(g, f)$ lies on $x+y=-4$.