JEE Main 2026 — 28 January, Evening Shift — Question 11

Area bounded between $P_{1}$ & $P_{2}$ is $\int_{-3}^{3}\left(\left(x^{2}+27\right)-\left(4 x^{2}\right)\right) d x$ (P.O.I. of $P_{1}$ & $P_{2}$ is $x= \pm 3$ )

$=2 \int_{0}^{3}\left(27-3 \mathrm{x}^{2}\right) \mathrm{dx}=2\left[27 \mathrm{x}-\mathrm{x}^{3}\right]_{0}^{3}$

$=2[81-27]=108$

∴ Area bounded between $\mathrm{P}_{1}$ & L is 18 sq. units (Area between $x^{2}=4$ ay & line $x=m y$ ) is $\frac{8 a^{2}}{3 m^{3}}$

∴ Area between $\mathrm{x}^{2}=\frac{\mathrm{y}}{4} \& \mathrm{x}=\frac{\mathrm{y}}{\alpha}$ is $\frac{8 \cdot\left(\frac{1}{16}\right)^{2}}{3 \cdot\left(\frac{1}{\alpha}\right)^{3}}=18$

$\Rightarrow \frac{\frac{8}{16.16}}{\frac{3}{\alpha^{3}}}=18 \Rightarrow \alpha^{3}=2^{6} .3^{3}$

$\Rightarrow \alpha=12$