JEE Main 2026 — Apr 4, Shift 2 — Mathematics solution

JEE Main2026Apr 4, Shift 2Mathematics

The area of the region bounded by the curves $x + 3 y^{2} = 0$ and $x + 4 y^{2} = 1$ is equal to:

A
$\frac{1}{3}$
B
$\frac{2}{3}$
C
$\frac{4}{3}$
D
$\frac{5}{3}$

Solution

Rewrite: $x = - 3 y^{2}$ (a leftward parabola through origin) and $x = 1 - 4 y^{2}$ (a leftward parabola through $(1, 0)$ ).

Intersections: $- 3 y^{2} = 1 - 4 y^{2} \Rightarrow y^{2} = 1 \Rightarrow y = \pm 1$ .

Between these $y$ -values, the curve $x = 1 - 4 y^{2}$ lies to the right of $x = - 3 y^{2}$ (test at $y = 0$ : $1 > 0$ ).

$A = \int_{- 1}^{1} [(1 - 4 y^{2}) - (- 3 y^{2})] d y = \int_{- 1}^{1} (1 - y^{2}) d y = 2 \int_{0}^{1} (1 - y^{2}) d y = 2 (1 - \frac{1}{3}) = \frac{4}{3}$ .

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