If the angle between the asymptotes is `30^(@)` then find its eccentricity.

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 asymptotes of the hyperbola
The standard form of a hyperbola is given by:

\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

The equations of the asymptotes for this hyperbola are:

\[ y = \pm \frac{b}{a} x \]

### Step 2: Use the formula for the angle between two lines
The angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) is given by:

\[ \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2} \]

For our asymptotes, the slopes are \(m_1 = \frac{b}{a}\) and \(m_2 = -\frac{b}{a}\). Thus, we have:

\[ \tan 30^\circ = \frac{\frac{b}{a} - \left(-\frac{b}{a}\right)}{1 + \frac{b}{a} \cdot \left(-\frac{b}{a}\right)} \]

### Step 3: Substitute the values
Substituting \(\tan 30^\circ = \frac{1}{\sqrt{3}}\):

\[ \frac{1}{\sqrt{3}} = \frac{\frac{b}{a} + \frac{b}{a}}{1 - \left(\frac{b}{a}\right)^2} \]

This simplifies to:

\[ \frac{1}{\sqrt{3}} = \frac{2\frac{b}{a}}{1 - \left(\frac{b}{a}\right)^2} \]

### Step 4: Let \(x = \frac{b}{a}\)
Let \(x = \frac{b}{a}\). Then the equation becomes:

\[ \frac{1}{\sqrt{3}} = \frac{2x}{1 - x^2} \]

### Step 5: Cross-multiply and rearrange
Cross-multiplying gives:

\[ 1 - x^2 = 2\sqrt{3}x \]

Rearranging this leads to:

\[ x^2 + 2\sqrt{3}x - 1 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

Here, \(a = 1\), \(b = 2\sqrt{3}\), and \(c = -1\):

\[ x = \frac{-2\sqrt{3} \pm \sqrt{(2\sqrt{3})^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]

Calculating the discriminant:

\[ (2\sqrt{3})^2 - 4(-1) = 12 + 4 = 16 \]

Thus,

\[ x = \frac{-2\sqrt{3} \pm 4}{2} \]

### Step 7: Find the values of \(x\)
This gives us two possible values:

\[ x = -\sqrt{3} + 2 \quad \text{and} \quad x = -\sqrt{3} - 2 \]

Since \(x = \frac{b}{a}\) must be positive, we take:

\[ x = 2 - \sqrt{3} \]

### Step 8: Calculate the eccentricity \(e\)
The eccentricity \(e\) of the hyperbola is given by:

\[ e = \sqrt{1 + \left(\frac{b}{a}\right)^2} = \sqrt{1 + x^2} \]

Substituting \(x = 2 - \sqrt{3}\):

\[ e = \sqrt{1 + (2 - \sqrt{3})^2} \]

Calculating \((2 - \sqrt{3})^2\):

\[ (2 - \sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3} \]

Thus,

\[ e = \sqrt{1 + 7 - 4\sqrt{3}} = \sqrt{8 - 4\sqrt{3}} \]

### Step 9: Final simplification
This can be factored out:

\[ e = \sqrt{4(2 - \sqrt{3})} = 2\sqrt{2 - \sqrt{3}} \]

### Final Answer
The eccentricity \(e\) of the hyperbola is:

\[ e = 2\sqrt{2 - \sqrt{3}} \]