detailed solution

Correct option is D

The equations of the curves are

    $y^4=x$ and $xy=k$

These two curves intersect at $\left(k^{4/5},k^{1/5}\right).$

Now,

     $y^4=x$ and $xy=k$

$\Rightarrow 4y^3\frac{dy}{dx}=1$ and $x\frac{dy}{dx}+y=0$

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

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{C_1}=\frac{1}{4k^{3/5}}$ and ${\left(\frac{dy}{dx}\right)}_{C_2}=-k^{-3/5}$

If the curves intersect at right angle, then

 $\frac{1}{4k^{3/5}}\times -k^{-3/5}=-1\Rightarrow k^{6/5}=\frac{1}{4}\Rightarrow k^6={\left(\frac{1}{4}\right)}^5\Rightarrow (4k{)}^6=4$

detailed solution

Correct answer is 4

The equations of the curves are

    $y^4=x$ and $xy=k$

These two curves intersect at $\left(k^{4/5},k^{1/5}\right).$

Now,

     $y^4=x$ and $xy=k$

$\Rightarrow 4y^3\frac{dy}{dx}=1$ and $x\frac{dy}{dx}+y=0$

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

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{C_1}=\frac{1}{4k^{3/5}}$ and ${\left(\frac{dy}{dx}\right)}_{C_2}=-k^{-3/5}$

If the curves intersect at right angle, then

 $\frac{1}{4k^{3/5}}\times -k^{-3/5}=-1\Rightarrow k^{6/5}=\frac{1}{4}\Rightarrow k^6={\left(\frac{1}{4}\right)}^5\Rightarrow (4k{)}^6=4$