If the curves x=y^4 and x y=k

$x=y^{4} x y=k$

for intersection $\mathrm{y}^{5}=\mathrm{k} \ldots(1)$

Also $x=y^{4}$

$\Rightarrow 1=4 y^{3} \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{1}{4 y^{3}}$

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

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

$\because$ Curve cut orthogonally

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

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

$\therefore$ from (1) $\mathrm{y}^{5}=\mathrm{k}$

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

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