Given, $x=y^4$and $xy=k$

By solving equations, we get  $y^5=k \dots \left(1\right)$

Also $x=y^4$

$\Rightarrow 1=4y^3\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{1}{4y^3}$

for $xy=k\Rightarrow x=\frac{k}{y}$

$\Rightarrow 1=-\frac{k}{y^2}·\frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{-y^2}{k}$

$\because$ Curve cut orthogonally

$\Rightarrow \frac{1}{4y^3}\times \left(\frac{-y^2}{k}\right)=-1$

$\Rightarrow y=\frac{1}{4k}$

$\therefore$ from $\left(1\right) y^5=k$

$\Rightarrow \frac{1}{{\left(4k\right)}^5}=k$

$\Rightarrow 4={\left(4k\right)}^6$