Find the condition that the curves $2x = y^2$ and $2xy = k$ intersect orthogonally.

(D) Given,the equations of the curves are $2x = y^2 \dots(i)$ and $2xy = k \dots(ii)$.
From $(ii)$,$y = \frac{k}{2x}$. Substituting this into $(i)$,we get $2x = (\frac{k}{2x})^2 \Rightarrow 2x = \frac{k^2}{4x^2} \Rightarrow 8x^3 = k^2 \Rightarrow x = \frac{k^{2/3}}{2^{1/3} \cdot 2^{2/3}} = \frac{k^{2/3}}{2}$.
Then $y^2 = 2x = k^{2/3} \Rightarrow y = k^{1/3}$.
The point of intersection is $(\frac{k^{2/3}}{2}, k^{1/3})$.
Differentiating $(i)$ with respect to $x$: $2 = 2y \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{y}$. Let $m_1 = \frac{1}{k^{1/3}}$.
Differentiating $(ii)$ with respect to $x$: $2y + 2x \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$.
At the point of intersection,$m_2 = -\frac{k^{1/3}}{k^{2/3}/2} = -\frac{2}{k^{1/3}}$.
Since the curves intersect orthogonally,$m_1 \cdot m_2 = -1$.
$\frac{1}{k^{1/3}} \cdot (-\frac{2}{k^{1/3}}) = -1 \Rightarrow -\frac{2}{k^{2/3}} = -1 \Rightarrow k^{2/3} = 2$.
Cubing both sides,$k^2 = 8$.