The angle between the asymptotes of a hyperbola is `30^(@).` The eccentricity of the hyperbola may be

Options

A. `sqrt3 pm 1`
B. `sqrt3 + 1`
C. `sqrt6 pm sqrt2`
D. `sqrt5 pm sqrt2`

Answer

To find the eccentricity of a hyperbola given that the angle between its asymptotes is \(30^\circ\), we can follow these steps:

### Step 1: Understand the relationship between the angle and the hyperbola parameters
The angle \(\theta\) between the asymptotes of a hyperbola is related to the ratio \( \frac{b}{a} \) by the formula:
\[ \tan\left(\frac{\theta}{2}\right) = \frac{b}{a} \]
Given that \(\theta = 30^\circ\), we have:
\[ \frac{\theta}{2} = 15^\circ \]

### Step 2: Calculate \(\tan(15^\circ)\)
Using the tangent value:
\[ \tan(15^\circ) = \tan\left(\frac{\pi}{12}\right) = 2 - \sqrt{3} \]
Thus, we can write:
\[ \frac{b}{a} = 2 - \sqrt{3} \]

### Step 3: Use the eccentricity formula
The eccentricity \(e\) of a hyperbola is given by the formula:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting \(\frac{b}{a} = 2 - \sqrt{3}\):
\[ \frac{b^2}{a^2} = (2 - \sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3} \]
Thus, we have:
\[ e = \sqrt{1 + (7 - 4\sqrt{3})} = \sqrt{8 - 4\sqrt{3}} \]

### Step 4: Simplify the expression for eccentricity
We can rewrite \(e\) as:
\[ e = \sqrt{4(2 - \sqrt{3})} = 2\sqrt{2 - \sqrt{3}} \]

### Final Result
The eccentricity of the hyperbola is:
\[ e = 2\sqrt{2 - \sqrt{3}} \]