Question

If the angle between the asymptotes of a hyperbola is 30° then its eccentricity is

• A. √5 - √2
• B. √6 - √3
• C. √5 - √3
• D. √6 - √2

Answer 1

The Correct Option is D

To solve the problem, we need to find the eccentricity of a hyperbola given that the angle between its asymptotes is 30 degrees.

1. Understanding the Hyperbola and Asymptotes:
Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
The equations of the asymptotes are $y = \pm \frac{b}{a} x$.
The angle between the asymptotes is $2 \theta$, where $\tan \theta = \frac{b}{a}$.

2. Using the Given Angle:
Given the angle between the asymptotes is $30^\circ$, we have $2\theta = 30^\circ$, so $\theta = 15^\circ$.
Thus, $\tan(15^\circ) = \frac{b}{a}$.

3. Evaluating $\tan(15^\circ)$:
We know $\tan(15^\circ) = 2 - \sqrt{3}$.
Therefore, $\frac{b}{a} = 2 - \sqrt{3}$.

4. Calculating the Eccentricity:
The eccentricity is $e = \sqrt{1 + \frac{b^2}{a^2}}$.
Substitute $\frac{b}{a} = 2 - \sqrt{3}$, so $e = \sqrt{1 + (2 - \sqrt{3})^2}$.
Compute $(2 - \sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3}$.
Thus, $e = \sqrt{1 + (7 - 4\sqrt{3})} = \sqrt{8 - 4\sqrt{3}} = \sqrt{6} - \sqrt{2}$.

Final Answer:
The eccentricity of the hyperbola is $\sqrt{6} - \sqrt{2}$.