JEE Main 2024 — 1 February, Shift 2 — Question 28

$\mathrm{ky}^{2}=2(\mathrm{y}-\mathrm{x})$

$2 y^{2}=k x$

Point of intersection $\rightarrow$

$\begin{gathered} k y^{2}=2\left(y-\frac{2 y^{2}}{k}\right) y=0 \end{gathered}$

$\quad k y=2\left(1-\frac{2 y}{k}\right)$

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

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

$A=\int_{0}^{\frac{2 k}{k^{2}+4}}\left(\left(y-\frac{k y^{2}}{2}\right)-\left(\frac{2 y^{2}}{k}\right)\right) \cdot d y =\frac{y^{2}}{2}-\left.\left(\frac{k}{2}+\frac{2}{k}\right) \cdot \frac{y^{3}}{3}\right|_{0} ^{\frac{2 k}{k^{2}+4}} =\left(\frac{2 k}{k^{2}+4}\right)^{2}\left[\frac{1}{2}-\frac{k^{2}+4}{2 k} \times \frac{1}{3} \times \frac{2 k}{k^{2}+4}\right]$

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

$A \cdot M \geq G \cdot M$

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

$\mathrm{k}+\frac{4}{\mathrm{k}} \geq 4$

Area is maximum when $k=\frac{4}{k}$

$\mathrm{k}=2,-2$

The sum of squares of all possible values of $k$ is $8$.