JEE Main 2024 — Area Under Curves Question with Solution

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

Finding the point of intersection,

$\Rightarrow k{y}^2=2\left(y-\frac{2y^2}{k}\right)$

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

$\Rightarrow ky+\frac{4y}{k}=2$

$\Rightarrow y=\frac{2}{k+\frac{4}{k}}$

$\Rightarrow y=\frac{2k}{k^2+4}$

So, the required area is given by,

$A=\underset{0}{\overset{\frac{2k}{k^2+4}}{\int }}\left(\left(y-\frac{k{y}^2}{2}\right)-\left(\frac{2y^2}{k}\right)\right)\cdot dy$

$\Rightarrow A={\left[\frac{y^2}{2}-\left(\frac{k}{2}+\frac{2}{k}\right)\cdot \frac{y^3}{3}\right]}_0^{\frac{2k}{k^2+4}}$

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

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

We know that, $AM\ge GM$

$\Rightarrow \frac{\left(k+\frac{4}{k}\right)}{2}\ge 2$

$\Rightarrow k+\frac{4}{k}\ge 4$

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

$\Rightarrow k=2,-2$

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