JEE Main 2026 (Online) 28th January Evening ShiftQuestion 1
Let $P_1:y=4x^2$ and $P_2:y=x^2+27$ be two parabolas. If the area of the bounded region enclosed between $P_1$ and $P_2$ is six times the area of the bounded region enclosed between the line $y=\alpha x$, $\alpha>0$ and $P_1$, then $\alpha$ is equal to :
Application of Integrals
Solution
$P_1\cap P_2$: $3x^2=27$, $x=\pm3$; area $=\int_{-3}^3(27-3x^2)dx=108$. Area between $y=\alpha x$ and $y=4x^2$ is $\frac{4(\alpha/4)^3}6=\frac{\alpha^3}{96}$. So $108=\frac{6\alpha^3}{96}$, $\alpha^3=1728$, $\alpha=12$.