Two parabolas with a common vertex and with axes along x -axis and...

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. $$\begin{aligned}&\therefore y²=3 x \Rightarrow(m x+c)^2=3 x \\ &\Rightarrow m² x²+(2 m c-3) x+c²=0 \end{aligned}$$As, the tangent touches at one point only So, $b^2-4 a c=0$$\begin{aligned}&\Rightarrow(2 m c-3)^2-4 m² c²=0 \\ &\Rightarrow 4 m² c²+9-12 m c-4 m² c²=0 \end{aligned}$$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} \Rightarrow4(x+y)+3=0 $