Given curve, $3 x^2-y^2-2 x+4 y=0$
Chord passes through $(1,-2)$ cut the given curve at $P$ and $Q$


Let slope of line $P Q$ is $m$.
Then, equation of line is
$$
\begin{aligned}
& y-y_1=m\left(x-x_1\right) \\
& \Rightarrow \quad y+2=m(x-1) \\
& y+2=m x-m \\
& \Rightarrow \quad \frac{m x-y}{2+m}=1 \\
&
\end{aligned}
$$

Use Eq. (ii) in Eq. (i) as follows,
$$
\begin{aligned}
& 3 x^2-y^2-2 x\left(\frac{m x-y}{2+m}\right)+4 y\left(\frac{m x-y}{2+m}\right)=0 \\
& \Rightarrow 3(2+m) x^2-(2+m) y^2-2 x(m x-y) \\
&+ 4 y(m x-y)=0 \\
& \Rightarrow x^2(6+3 m-2 m)-y^2(2+m+4)+x y \\
&(4 m+2)=0
\end{aligned}
$$

General equation of curve is
$$
a x^2+2 h x y+b y^2+2 g x+2 f y+c=0
$$

Compare Eqs. (iii) and (iv),
$$
a=6+m, b=-6-m
$$

This gives, $a+b=0$
$\Rightarrow$ Angle subtended by these two lines at origin is $90^{\circ}$.