\(x^2 - 4y^2 = 1 \Rightarrow \frac{x^2}{1} - \frac{y^2}{1/4} = 1\), so \(a^2 = 1\) and \(b^2 = \frac{1}{4}\).

For a hyperbola, \(c^2 = a^2 + b^2 = 1 + \frac{1}{4} = \frac{5}{4}\), so \(c = \frac{\sqrt{5}}{2}\).

The foci are at \((\pm c, 0)\), so the distance between them = \(2c = \sqrt{5}\).

Answer: (B) \(\sqrt{5}\)