Minimum distance between the parabolas $y^2-4x-8y+40=0$  and $x^2-8x-4y+40=0$ is 

Since two parabolas are symmetrical about y=x

Slope of tangent is equal to 1

Differentiating $x^2-8x-4y+40=0$ with respect to x. 

we get $2x-8-4\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{x-4}{2}=1$

(slope of tangent )so, x=6 and y=7

$\therefore A=\left(6,7\right),B=\left(7,6\right)$

So, minimum distance is  $AB=\sqrt{2}$