The minimum distance between the parabolas ${\mathrm{y}}^2-4\mathrm{x}-8\mathrm{y}+40=0$ and 

${\mathrm{x}}^2-8\mathrm{x}-4\mathrm{y}+40=0$ is _______

$$\begin{gathered} {\mathrm{y}}^2-4\mathrm{x}-8\mathrm{y}+40=0. -\left(1\right) {\mathrm{x}}^2-8\mathrm{x}-4\mathrm{y}+40=0 -\left(2\right) \\ The parabolas \left(1\right) and \left(2\right) are symmetric about y=x \\ ⟹y=\left(1\right)x \\ \therefore Slope=1 \\ Minimum dis\tan ce (points are common \tan gent and common normal) \end{gathered}$$

Let P(x, y) be point on  (1)

$$\begin{gathered} 2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}-4-8\frac{\mathrm{dy}}{\mathrm{dx}}=0 \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4}{2\mathrm{y}-8}=\frac{2}{\mathrm{y}-4}=1 \\ \mathrm{y}-4=2 \\ \mathrm{y}=6 \end{gathered}$$

  $$\begin{gathered} from \left(1\right) \Rightarrow 36-4x-48+40=0 \\ \Rightarrow 4\mathrm{x}=28 \\ \Rightarrow \mathrm{x}=7 \\ So \mathrm{P}(7, 6) \\ point on second parabola \mathrm{Q}=(6, 7) \sin ce it is symmetrical about y=x \end{gathered}$$

Minimum  distance $\mathrm{PQ}=\sqrt{1+1}=\sqrt{2}$