As origin is the only common point to $x-$ axis and $y$-axis, so, origin is the common vertex Let the equation of two of parabolas be $y^2=4 a x$ and $x^2=4 b y$
Now latus rectum of both parabolas $=3$
$$
\begin{aligned}
&\therefore 4 a=4 b=3 \\
&\Rightarrow a=b=\frac{3}{4}
\end{aligned}
$$
$\therefore$ Two parabolas are $y^2=3 x$ and $x^2=3 y$ Suppose $y=m x+c$ is the common tangent.
$$
\begin{aligned}
&\therefore y^2=3 x \Rightarrow(m x+c)^2=3 x \\
&\Rightarrow m^2 x^2+(2 m c-3) x+c^2=0
\end{aligned}
$$
As, the tangent touches at one point only
So, $b^2-4 a c=0$
$$
\begin{aligned}
&\Rightarrow(2 m c-3)^2-4 m^2 c^2=0 \\
&\Rightarrow 4 m^2 c^2+9-12 m c-4 m^2 c^2=0
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow c=\frac{9}{12 m}=\frac{3}{4 m} \\
&\therefore x^2=3 y \Rightarrow x^2=3(m x+c) \\
&\Rightarrow x^2-3 m x-3 c=0
\end{aligned}
$$
Form (i) and ( ii)
$$
\begin{aligned}
&m^2=\frac{-4 c}{3}=\frac{-4}{3}\left(\frac{3}{4 m}\right) \\
&\Rightarrow m^3=-1 \Rightarrow m=-1 \Rightarrow c=\frac{-3}{4}
\end{aligned}
$$
Hence, $y=m x+c=-x-\frac{3}{4}$
$$
\Rightarrow 4(x+y)+3=0
$$