If the curves x = y ^ 4 and xy = k cut at right angles, then (4k) ^...

Given, $x = {y}^{4}$and $xy = k$

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

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

$⇒ 1 = 4{y}^{3}\dfrac{dy}{dx} ⇒ \dfrac{dy}{dx} = \dfrac{1}{4{y}^{3}}$

for $xy = k ⇒ x = \dfrac{k}{y}$

$⇒ 1 = - \dfrac{k}{{y}^{2}} \cdot \dfrac{dy}{dx}$

$⇒ \dfrac{dy}{dx} = \dfrac{- {y}^{2}}{k}$

$∵$ Curve cut orthogonally

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

$⇒ y = \dfrac{1}{4k}$

$∴$ from \left(1\right){y}^{5} = k$</p><p>$⇒ \dfrac{1}{{\left(4k\right)}^{5}} = k$</p><p>$⇒ 4 = {\left(4k\right)}^{6}$