If the curves x = y⁴ and xy = k cut at right angles, then (4k)⁶ is equal to __________.

$x = {y^4}$ and $xy = k$<br><br>for intersection ${y^5} = k$ ..... (1)<br><br>Also $x = {y^4}$ <br><br>$$ \Rightarrow 1 = 4{y^3}{{dy} \over {dx}} \Rightarrow {{dy} \over {dx}} = {1 \over {4{y^3}}}$$<br><br>for $xy = k \Rightarrow x = {k \over y}$<br><br>$\Rightarrow 1 = - {k \over {{y^2}}}.{{dy} \over {dx}}$<br><br>$\Rightarrow {{dy} \over {dx}} = {{ - {y^2}} \over k}$<br><br>$\because$ Curve cut orthogonally<br><br>$$ \Rightarrow {1 \over {4{y^3}}} \times \left( {{{ - {y^2}} \over k}} \right) = - 1$$<br><br>$\Rightarrow y = {1 \over {4k}}$<br><br>$\therefore$ from (1), ${y^5} = k$<br><br>$\Rightarrow {1 \over {{{(4k)}^5}}} = k$<br><br>$\Rightarrow 4 = {(4k)^6}$