JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The given equation of the hyperbola is x^2 a^2 - y^2 b^2 = 1. Since it passes through (6, 4 3 ), we have: 36 a^2 - 48 b^2 = 1 The eccentricity e satisfies 15(e^2 + 1) = 34e. 15e^2 - 34e + 15 = 0 (3e - 5)(5e - 3) = 0 Since the eccentricity of a hyperbola is e > 1, we get e = 5 3 . We know that b^2 = a^2(e^2 - 1). b^2 = a^2 ( 25 9 - 1 ) = 16 9 a^2 Substituting b^2 into the first equation: 36 a^2 - 48 16 9 a^2 = 1 36 a^2 - 27 a^2 = 1 9 a^2 = 1 a^2 = 9 Then, b^2 = 16 9 9 = 16. The second hyperbola is given by x^2 b^2 - y^2 2(a^2+1) = 1. Substituting the values of a^2 and b^2: x^2 16 - y^2 2(9+1) = 1 x^2 16 - y^2 20 = 1 Here, A^2 = 16 A = 4 and B^2 = 20. The length of the latus rectum is 2B^2 A . Length of latus rectum = 2 20 4 = 10. Answer: 10