JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Given the equation of the ellipse is x^2 a^2 + y^2 b^2 = 1 with a The eccentricity is given by e = 5 3 . Since a Substituting the value of e: a^2 = b^2 (1 - 5 9 ) = 4 9 b^2 The ellipse passes through the point (4, 3), so substituting x = 4 and y = 3 into the equation of the ellipse gives: 16 a^2 + 9 b^2 = 1 Substituting a^2 = 4 9 b^2 into the above equation: 16 4 9 b^2 + 9 b^2 = 1 36 b^2 + 9 b^2 = 1 45 b^2 = 1 b^2 = 45 Now, finding a^2: a^2 = 4 9 45 = 20 For an ellipse with a Substituting the values of a^2 and b = 45 = 3 5 : Length of latus rectum = 2 20 3 5 = 40 3 5 = 8 5 3 Answer: 8 5 3