NDA Mathematics Question (2025) — Solution

Step-by-step solution

The given equation of the hyperbola is x^2 - 4y^2 = 1. This can be rewritten in the standard form as x^2 1 - y^2 1/4 = 1. Comparing with x^2 a^2 - y^2 b^2 = 1, we get a^2 = 1 and b^2 = 1 4 . The eccentricity e of the hyperbola is given by e = 1 + b^2 a^2 . e = 1 + 1/4 1 = 5 4 = 5 2 The distance between the foci of the hyperbola is 2ae. 2ae = 2 1 5 2 = 5 Answer: 5