$x-2y+4=0$ is a common tangent to ${\mathrm{y}}^2=4\mathrm{x} \mathrm{\&} \frac{{\mathrm{x}}^2}{4}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ Then the value of b and the other common tangent are given by:

 $x-2y+4=0$is a tangent to  $\frac{{\mathrm{x}}^2}{4}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$

$\begin{array}{l}\Rightarrow {\mathrm{a}}^2{\mathrm{l}}^2+{\mathrm{b}}^2{\mathrm{m}}^2={\mathrm{n}}^2\\ \Rightarrow 4+{\mathrm{b}}^2\left(4\right)=16\Rightarrow {\mathrm{b}}^2=3\end{array}$

The other tangent must be reflection of x-2y+4 = 0 in the x-axis since both curves are symmetrical about x-axis.

$x+2y+4=0$ is the other common tangent.