The distance of the point $(6,-2\sqrt{2})$  from the common tangent $\mathrm{y}=\mathrm{mx}+\mathrm{c},\hspace{0.17em}\mathrm{m}>0,$  of the curves $\mathrm{x}=2{\mathrm{y}}^2$  and $\mathrm{x}=1+{\mathrm{y}}^2$  is:

$\begin{array}{l}{\mathrm{y}}^2=\frac{\mathrm{x}}{2} equation of\text{ tangent is }\mathrm{y}=\mathrm{mx}+\frac{1}{8\mathrm{m}}\\ {\mathrm{y}}^2=\mathrm{x}-1 equation of \mathrm{tangent} is \mathrm{y}=\mathrm{m}(\mathrm{x}-1)+\frac{1}{4\mathrm{m}}\end{array}$
$$\begin{gathered} \begin{array}{l}\therefore \frac{1}{8\mathrm{m}}=-\mathrm{m}+\frac{1}{4\mathrm{m}} \\ \begin{array}{r}\Rightarrow 1=-8m^2+2\\ \because m>0\Rightarrow m=\frac{1}{2\sqrt{2}}\end{array}\\ \therefore \Rightarrow \text{ Common tangent is }y=\frac{x}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}\\ \Rightarrow x-2\sqrt{2}y+1=0\end{array} \end{gathered}$$
$d=\frac{\left|6+8+1\right|}{\sqrt{1+8}}=\frac{15}{3}=5$