The area of the region $A = \{(x, y) : 4x^2 + y^2 \leq 8 \text{ and } y^2 \leq 4x\}$ is :

(D) $\frac{\pi}{2} + \frac{1}{3}$

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Area Under Curves

The area of the region

A = {(x, y) : 4 x^{2} + y^{2} \leq 8 and y^{2} \leq 4 x}

is :

Answer: (B)

The ellipse

4 x^{2} + y^{2} = 8

and the parabola

y^{2} = 4 x

meet where

4 x^{2} + 4 x = 8

, i.e.

x^{2} + x - 2 = 0

, giving

x = 1

(with

y = \pm 2

). The region is symmetric about the

x

-axis; for

0 \leq x \leq 1

it is bounded by the parabola and for

1 \leq x \leq 2

by the ellipse, so

Area = \int_{0}^{1} 4 x d x + \int_{1}^{2} 2 8 - 4 x^{2} d x = \frac{8}{3} + 4 \int_{1}^{2} 2 - x^{2} d x .

Using

\int 2 - x^{2} d x = \frac{x}{2} 2 - x^{2} + sin^{- 1} \frac{x}{2}

, the second integral equals

4 [\frac{π}{2} - (\frac{1}{2} + \frac{π}{4})] = π - 2

. Hence the area of the region is

\frac{8}{3} + π - 2 = π + \frac{2}{3}

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Single correct option