Let \(P_{1}: y=4 x^{2}\) and \(P_{2}: y=x^{2}+27\) be two parabolas. If the area of the bounded region enclosed between \(\mathrm{P}_{1}\) and \(\mathrm{P}_{2}\) is six times the area of the bounded region enclosed between the line \(\mathrm{y}=\alpha \mathrm{x}, \alpha>0\) and \(\mathrm{P}_{1}\), then \(\alpha\) is equal to:

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Area Under Curves

Let

P_{1} : y = 4 x^{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 = α x, α > 0

and

P_{1}

, then

α

is equal to:

Answer: (C)

The parabolas meet where

4 x^{2} = x^{2} + 27

, i.e.

x = \pm 3

, so the region between them has area

\int_{- 3}^{3} [(x^{2} + 27) - 4 x^{2}] d x = 2 \int_{0}^{3} (27 - 3 x^{2}) d x = 2 [27 x - x^{3}]_{0}^{3} = 2 (81 - 27) = 108.

The line

y = α x

meets

y = 4 x^{2}

x = 0

and

x = \frac{α}{4}

, and the region between them has area

\int_{0}^{α /4} (α x - 4 x^{2}) d x = [\frac{α x ^{2}}{2} - \frac{4 x ^{3}}{3}]_{0}^{α /4} = \frac{α ^{3}}{32} - \frac{α ^{3}}{48} = \frac{α ^{3}}{96} .

Setting the first area equal to six times the second,

108 = 6 \cdot \frac{α ^{3}}{96} = \frac{α ^{3}}{16}

, so

α^{3} = 1728

and

α = 12

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option