Q215 / 270 · NDA 2025 (Phase 2) · Mathematics
What is the distance between the foci of the hyperbola $x^2 - 4y^2 = 1$?
- A$\sqrt{3}$
- Correct answerB$\sqrt{5}$
- C$2\sqrt{3}$
- D$2\sqrt{5}$
Solution
The given equation of the hyperbola is $x^2 - 4y^2 = 1$.
This can be rewritten in the standard form as $\dfrac{x^2}{1} - \dfrac{y^2}{1/4} = 1$.
Comparing with $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$, we get $a^2 = 1$ and $b^2 = \dfrac{1}{4}$.
The eccentricity $e$ of the hyperbola is given by $e = \sqrt{1 + \dfrac{b^2}{a^2}}$.
$e = \sqrt{1 + \dfrac{1/4}{1}} = \sqrt{\dfrac{5}{4}} = \dfrac{\sqrt{5}}{2}$
The distance between the foci of the hyperbola is $2ae$.
$2ae = 2 \times 1 \times \dfrac{\sqrt{5}}{2} = \sqrt{5}$
Answer: $\sqrt{5}$
This can be rewritten in the standard form as $\dfrac{x^2}{1} - \dfrac{y^2}{1/4} = 1$.
Comparing with $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$, we get $a^2 = 1$ and $b^2 = \dfrac{1}{4}$.
The eccentricity $e$ of the hyperbola is given by $e = \sqrt{1 + \dfrac{b^2}{a^2}}$.
$e = \sqrt{1 + \dfrac{1/4}{1}} = \sqrt{\dfrac{5}{4}} = \dfrac{\sqrt{5}}{2}$
The distance between the foci of the hyperbola is $2ae$.
$2ae = 2 \times 1 \times \dfrac{\sqrt{5}}{2} = \sqrt{5}$
Answer: $\sqrt{5}$
Hence, the correct answer is option: 'B'.