Given $y^2=x$ and $xy=k $ are orthogonal

Step 1:

Let the curves intersect at $ (x_1,y_1) $

Therefore ${y_1}^2=x_1$----(i) and $x_1y_1=k$-----(ii)

Substitute for $y_1$ from (ii) in (i)

$\large\frac{k^2}{{x_1}^2}$$=x_1=>{x_1}^3=k^2$-----(iii)

Step 2:

The slopes of the tangents of the curves $(m_1,m_2)$ at $(x_1,y_1)$ is $\large\frac{dy}{dx{(x_1y_1)}}$ for each curve

$y^2=x$

$2y \large\frac{dy}{dx}$$=1$

$\large\frac{dy}{dx}=\frac{1}{2y}$

Therefore $m_1=\large\frac{1}{2y_1}=\frac{x_1}{2k}$ (from (ii))

$xy=k$

$y+x\large\frac{dy}{dx}$$=0=> \large\frac{dy}{dx}=\frac{-y}{x}$

$m_2=\large\frac{-y_1}{x_1}=\frac{-k}{{x_1}^2}$ (from (ii))

Step 3:

The curves are orthogonal

Therefore $m_1m_2=-1 =>\large\frac{x_1}{2k} \times \frac{-k}{{x_1}^2}$$=-1=>x_1=\large\frac{1}{2}$

Step 4:

Substitute $x_1=\large\frac{1}{2}$ in (iii)

$\large\frac{1}{8}$$=k^2 =>8h^2=1$