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 $=3$
$\therefore 4 a=4 b=3$
$\Rightarrow a=b=\frac{3}{4}$
$\therefore$ Two parabolas are $y^{2}=3 x$ and $x^{2}=3 y$
Suppose $y=m x+c$ is the common tangent.
$\therefore$ $y^{2}=3 x \Rightarrow(m x+c)^{2}=3 x$
$\Rightarrow m^{2} x^{2}+(2 m c-3) x+c^{2}=0$
As, the tangent touches at one point only $\mathrm{So}, b^{2}-4 a c=0$
$\Rightarrow(2 m c-3)^{2}-4 m^{2} c^{2}=0$
$\Rightarrow 4 m^{2} c^{2}+9-12 m c-4 m^{2} c^{2}=0$
$\Rightarrow c=\frac{9}{12 m}=\frac{3}{4 m}$ ......$(i)$
$\therefore x^{2}=3 y \Rightarrow x^{2}=3(m x+c)$
$\Rightarrow x^{2}-3 m x-3 c=0$
Again, $b^{2}-4 a c=0$
$\Rightarrow 9 m^{2}-4(1)(-3 c)=0$
$\Rightarrow 9 m^{2}=-12 c$ .......$(ii)$
Form $(i)$ and $(ii)$
$m^{2}=\frac{-4 c}{3}=\frac{-4}{3}\left(\frac{-3}{4 m}\right)$
$\Rightarrow m^{3}=-1 \Rightarrow m=-1 \Rightarrow c=\frac{-3}{4}$
Hence, $y=m x+c=-x-\frac{3}{4}$
$\Rightarrow 4(x+y)+3=0$
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