The equation of the common tangent to the  parabolas  $y^2=4ax$ and $x^2=4$ by is 

The equation of any tangent to $y^2=4ax$ is 

$y=mx+\begin{array}{r}a\\ m\end{array}$                                    (i) 

If it is a tangent to the parabola $x^2=4$ $by$, then the equation

$x^2=4b\left(mx+\frac{a}{m}\right)$  must have equal roots.

$\Rightarrow$  $x^2-4bmx-\frac{4ab}{m}=0$ must have equal roots

$\Rightarrow 16b^2{m}^2+\frac{16ab}{m}=0$                            [On equating Disc. to zero]

 $m^3=-\frac{a}{b}\Rightarrow m=-{\left(\frac{a}{b}\right)}^{\frac{1}{3}}$

Putting the value of m in (i), we get 

$y=-{\left(\frac{a}{b}\right)}^{1/3}x-a^{2/3}{b}^{1/3}\Rightarrow a^{1/3}x+b^{1/3}y+a^{2/3}{b}^{2/3}=0$

This is the equation of the common tangent

$ALITER-$  The equation of any tangent to  $y^2=4ax$ is

   $y=mx+\frac{a}{m}$

$\Rightarrow x=\frac{y}{m}-\frac{a}{m^2}\Rightarrow x=m^{\mathrm{\prime }}y+(-a)m^{\mathrm{\prime }2},$ where $m^{\mathrm{\prime }}=+\frac{1}{m}$

If it touches the parabola $x^2=4by$, then we must have

$\begin{array}{l}-a{m}^{\mathrm{\prime }2}=\frac{b}{m^{\mathrm{\prime }}}\\ \Rightarrow m^{\mathrm{\prime }3}=-\frac{b}{a}\Rightarrow \frac{1}{m^3}=-\frac{b}{a}\Rightarrow m^3=-\frac{a}{b}\Rightarrow m=-{\left(\frac{a}{b}\right)}^{1/3}\end{array}$

Substituting the value of m in (i), we obtain the equation of the 

required common tangent as 

 $a^{1/3}x+b^{1/3}y+a^{2/3}{b}^{2/3}=0$