The shortest distance between the parabolas $2y^2=2x-1$ and $2x^2=2y-1$ is

Given parabolas are symmetrical about the line  $y=x$
The shortest distance occurs along the common normal which is perpendicular to the line  $y=x$
The tangent at point A on
                           
$2y^2=2x-1$ is parallel to $y=x$  
  
$4y\frac{dy}{dx}=2$   $\Rightarrow \frac{dy}{dx}=\frac{1}{2y}=1$$y=\frac{1}{2}$ and  $x=\frac{3}{4}$
$A(\frac{3}{4},\frac{1}{2})$   Then the coordinates of B on $2x^2=2y-1$
is  $B=(\frac{1}{2},\frac{3}{4})$
Shortest distance  $AB=\sqrt{{(\frac{3}{4}-\frac{1}{2})}^2+{(\frac{1}{2}-\frac{3}{4})}^2}=\frac{1}{2\sqrt{2}}$