Two parabolas with a common vertex and with axes along the x -axis and y -axis respectively, intersect each…

Equation of two parabola are $y^2=3x$ and $x^2=3y$.

Let equation of tangent to $y^2=3x$ is $y=mx+\frac{3}{4m}$ is also tangent to $x^2=3y$

$\Rightarrow x^2=3mx+\frac{9}{4m}$

$\Rightarrow 4m{x}^2-12m^{2}x-9=0$ have equal roots

$\Rightarrow D=0$

$\Rightarrow 144m^4=4\left(4m\right)(-9)$

$\Rightarrow m^4+m=0\Rightarrow m=-1$

Hence, common tangent is $y=-x-\frac{3}{4}$

$\Rightarrow 4\left(x+y\right)+3=0$