JEE Main 2026 Jan 28 Shift 2, Mathematics Q11: Area as Definite Integral

Area between $P_1$ and $P_2$: Intersections at $4x^2=x^2+27\Rightarrow x=\pm 3$. By symmetry,

$A_1=2{\int }_0^3\left((x^2+27)-4x^2\right)dx=2{\int }_0^3(27-3x^2)dx=2[27x-x^3{]}_0^3=2(81-27)=108$.

Area between $y=\alpha x$ and $P_1:y=4x^2$: Rewriting $P_1$ as $x^2=\frac{y}{4}$. Using the standard result that the area between the parabola $x^2=4ay$ and the chord $y=mx$ is $\frac{8a^2}{3m^3}$, with $4a=\frac{1}{4}$ so $a=\frac{1}{16}$, and slope $m=\alpha$:

$A_2=\frac{8(1/16{)}^2}{3{\alpha }^3}=\frac{1}{96{\alpha }^3}$. Hmm — let's compute directly instead. Intersections: $4x^2=\alpha x\Rightarrow x=0$ or $x=\alpha /4$.

$A_2={\int }_0^{\alpha /4}(\alpha x-4x^2)dx=\frac{\alpha \cdot (\alpha /4{)}^2}{2}-\frac{4(\alpha /4{)}^3}{3}=\frac{{\alpha }^3}{32}-\frac{{\alpha }^3}{48}=\frac{{\alpha }^3}{96}$.

Condition: $108=6\cdot \frac{{\alpha }^3}{96}\Rightarrow {\alpha }^3=\frac{108\cdot 96}{6}=1728\Rightarrow \alpha =12$.