The sum of squares of all possible values of k , for which area of...

Given: $k{y}^{2} = 2\left(y - x\right), 2{y}^{2} = kx$

Finding the point of intersection,

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

$⇒ y = 0$ and $ky = 2\left(1 - \dfrac{2y}{k}\right)$

$⇒ ky + \dfrac{4y}{k} = 2$

$⇒ y = \dfrac{2}{k + \dfrac{4}{k}}$

$⇒ y = \dfrac{2k}{{k}^{2} + 4}$

So, the required area is given by,

$A = \mathop{∫}_{0}^{\dfrac{2k}{{k}^{2} + 4}}\left(\left(y - \dfrac{k{y}^{2}}{2}\right) - \left(\dfrac{2{y}^{2}}{k}\right)\right) ⋅ dy$

$⇒ A = {\left[\dfrac{{y}^{2}}{2} - \left(\dfrac{k}{2} + \dfrac{2}{k}\right) ⋅ \dfrac{{y}^{3}}{3}\right]}_{0}^{\dfrac{2k}{{k}^{2} + 4}}$

$⇒ A = {\left(\dfrac{2k}{{k}^{2} + 4}\right)}^{2}\left[\dfrac{1}{2} - \dfrac{{k}^{2} + 4}{2k} \times \dfrac{1}{3} \times \dfrac{2k}{{k}^{2} + 4}\right]$

$⇒ A = \dfrac{1}{6} \times 4 \times {\left(\dfrac{1}{k + \dfrac{4}{k}}\right)}^{2}$

We know that, $AM \ge GM$

$⇒ \dfrac{\left(k + \dfrac{4}{k}\right)}{2} \ge 2$

$⇒ k + \dfrac{4}{k} \ge 4$

So, area is maximum when $k = \dfrac{4}{k}$.

$⇒ k = 2, - 2$

Hence, the sum of squares of all possible values of $k$ is $8$.