JEE Advanced Mathematics Question (2026) — Solution

Step-by-step solution

The given ellipses are E_1: x^2 + 4y^2 = 1 and E_2: 4x^2 + y^2 = 1. By symmetry, the common region is symmetric about both the coordinate axes and the lines y = x and y = -x. The total area is 8 times the area of the region in the first quadrant bounded by = 0 and = 4 . In polar coordinates, substituting x = r and y = r into the equation of E_2 (which is the inner boundary for 0 4 ), we get: 4r^2 ^2 + r^2 ^2 = 1 r^2 = 1 4 ^2 + ^2 The area of this sector is given by: A = 1 2 _ 0 ^ 4 r^2 d = 1 2 _ 0 ^ 4 1 4 ^2 + ^2 d Dividing the numerator and the denominator by ^2 : A = 1 2 _ 0 ^ 4 ^2 4 + ^2 d Substituting t = , we have dt = ^2 d . The limits change from 0 to 1: A = 1 2 _ 0 ^ 1 dt 4 + t^2 = 1 2 [ 1 2 ( t 2 ) ]_ 0 ^ 1 = 1 4 ( 1 2 ) The total area of the common region is: = 8A = 8 1 4 ( 1 2 ) = 2 ( 1 2 ) We need to find the value of : = (2 ( 1 2 ) ) Let = ( 1 2 ), which implies = 1 2 . Using the double angle formula for tangent: (2 ) = 2 1 - ^2 = 2 ( 1 2 ) 1 - ( 1 2 )^2 = 1 1 - 1 4 = 4 3 Therefore, = 1 (2 ) = 3 4 . Answer: 3/4