If the curves $2x=y^2$ and  $2xy=k$, intersect orthogonally, then the value of $k^2\text{\hspace{0.17em}=}$

Given equations of curves are
  $2x=y^2\mathrm{...}\left(i\right)$
And   $2xy=k\mathrm{...}\left(ii\right)$
Since curves intersect orthogonally,
${\left(\frac{dy}{dx}\right)}_1{\left(\frac{dy}{dx}\right)}_2=\left(\frac{1}{y}\right)(-\frac{y}{x})=-1$ or  $x=1$
So, from (i),  $y=\pm \sqrt{2}$
Putting these values in equation (ii), we get 
$k=\pm 2\sqrt{2}$

$\therefore k^2=8$