JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The equations of the given curves are x = -3y^2 and x = 1 - 4y^2. To find the points of intersection, equate the values of x: -3y^2 = 1 - 4y^2 y^2 = 1 y = 1 The region is bounded between y = -1 and y = 1. In this interval, 1 - 4y^2 -3y^2. The required area A is given by: A = _ -1 ^ 1 ( (1 - 4y^2) - (-3y^2) ) dy A = _ -1 ^ 1 (1 - y^2) dy Since 1 - y^2 is an even function: A = 2 _ 0 ^ 1 (1 - y^2) dy A = 2 [ y - y^3 3 ]_ 0 ^ 1 A = 2 ( 1 - 1 3 ) = 2 ( 2 3 ) = 4 3 Answer: 4 3