If the area enclosed between the curves y = k x ^ 2 and x = k y ^ 2...

The given curves are, $y = k{x}^{2}, x = k{y}^{2}$

To find the point of intersection of the curves, put $y$ from the first curve into second, to get

$x = k\left({k}^{2}{x}^{4}\right)$

$⇒ x = 0 or${x}^{3} = {\left(\dfrac{1}{k}\right)}^{3} ⇒ x = \dfrac{1}{k}$</p><p>Hence,$y = 0$or$y = k{\left(\dfrac{1}{k}\right)}^{2} = \dfrac{1}{k}$</p><p>Therefore, point of intersection is \left(\dfrac{1}{k}, \dfrac{1}{k}\right) .$

Hence, the required area is ${∫}_{0}^{\dfrac{1}{k}}\left({y}_{2} - {y}_{1}\right)dx = 1$

$⇒ {∫}_{0}^{\dfrac{1}{k}}\left(\sqrt{\dfrac{x}{k}} - k{x}^{2}\right)dx = 1$

$⇒ {\left(\dfrac{1}{\sqrt{k}}\dfrac{{x}^{3 / 2}}{3 / 2} - \dfrac{k{x}^{3}}{3}\right)}_{0}^{\dfrac{1}{k}} = 1$

$⇒ \dfrac{2}{3{k}^{2}} - \dfrac{1}{3{k}^{2}} = 1$

$⇒ {k}^{2} = \dfrac{1}{3}$

$⇒ k = \pm \sqrt{\dfrac{1}{3}}$

But, given $k > 0$

So, $k = \dfrac{1}{\sqrt{3}} .$