Explanation: Given curves, ${\mathit{y}}^2=4\mathit{x}$ and $\mathit{x}\mathit{y}=\mathit{c}$ cuts orthogonally.
Let they intersect at $\left({\mathit{x}}_1,{\mathit{y}}_1\right)$.
Now,
$\therefore 2\mathit{y}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=4$
$\Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{2}{\mathit{y}}$
$\Rightarrow {\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}|}_{\left({\mathit{x}}_1,{\mathit{y}}_1\right)}=\frac{2}{{\mathit{y}}_1}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(\mathit{i}\right)$

$\therefore \mathit{x}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}+\mathit{y}=0$
$\therefore \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{-\mathit{y}}{\mathit{x}}$
$\Rightarrow {\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}|}_{\left({\mathit{x}}_1,{\mathit{y}}_1\right)}=-\frac{{\mathit{y}}_1}{{\mathit{x}}_1}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(\mathit{i}\mathit{i}\right)$
From eqs. (i) and (ii)
$\frac{2}{{\mathit{y}}_1}\times \left(\frac{-{\mathit{y}}_1}{{\mathit{x}}_1}\right)=-1\left[\because {\mathit{m}}_1{\mathit{m}}_2=-1\right]$
$\Rightarrow {\mathit{x}}_1=2$

$\Rightarrow {\mathit{y}}_1^2=4\left(2\right)=8$
${\mathit{y}}_1=2\sqrt{2}$
Now, put value of ${\mathit{x}}_1$ and ${\mathit{y}}_1$ in ${\mathit{x}}_1{\mathit{y}}_1=\mathit{c}$, we get
$\mathit{c}=2\left(2\sqrt{2}\right)=4\sqrt{2}$