JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The given equations are: Hyperbola: 16x^2 - 9y^2 = 144 x^2 9 - y^2 16 = 1 Line: 8x - 3y = 24 y = 8 3 (x - 3) To find the points of intersection, substitute y from the line equation into the hyperbola equation: 16x^2 - 9 ( 8 3 (x - 3) )^2 = 144 16x^2 - 64(x^2 - 6x + 9) = 144 x^2 - 4(x^2 - 6x + 9) = 9 -3x^2 + 24x - 45 = 0 x^2 - 8x + 15 = 0 (x - 3)(x - 5) = 0 x = 3, x = 5 For x [3, 5], the upper curve is the hyperbola y = 4 3 x^2 - 9 and the lower curve is the line y = 8 3 (x - 3). The area A of the bounded region is: A = _ 3 ^ 5 ( 4 3 x^2 - 9 - 8 3 (x - 3) ) dx Using the standard integral formula x^2 - a^2 dx = x 2 x^2 - a^2 - a^2 2 |x + x^2 - a^2 |: _ 3 ^ 5 x^2 - 9 dx = [ x 2 x^2 - 9 - 9 2 |x + x^2 - 9 | ]_ 3 ^ 5 = ( 5 2 (4) - 9 2 9 ) - ( 0 - 9 2 3 ) = 10 - 9 3 + 9 2 3 = 10 - 9 2 3 For the linear part: _ 3 ^ 5 (x - 3) dx = [ (x - 3)^2 2 ]_ 3 ^ 5 = 4 2 - 0 = 2 Substituting these back into the area equation: A = 4 3 ( 10 - 9 2 3 ) - 8 3 (2) A = 40 3 - 6 3 - 16 3 A = 8 - 6 3 We need to find the value of 3(A + 6 _e 3): A + 6 3 = 8 3(A + 6 3) = 3 8 = 24 Answer: 24