If the latus rectum of a hyperbola forms an equilateral triangle with the vertex at the center of the hyperbola ,then find the eccentricity of the hyperbola.

Explanation

To solve the problem step by step, we will analyze the relationship between the latus rectum of the hyperbola and the equilateral triangle formed with the vertex at the center of the hyperbola.

### Step-by-Step Solution:

1. **Understanding the Hyperbola and Latus Rectum**:
- The latus rectum of a hyperbola is a line segment perpendicular to the transverse axis through a focus of the hyperbola. The length of the latus rectum is given by the formula \( \frac{2b^2}{a} \), where \( a \) and \( b \) are the semi-major and semi-minor axes of the hyperbola, respectively.

2. **Equilateral Triangle Formation**:
- Given that the latus rectum forms an equilateral triangle with the vertex at the center of the hyperbola, we can denote the center of the hyperbola as point \( O \) and the endpoints of the latus rectum as points \( A \) and \( B \). The triangle \( OAB \) is equilateral.

3. **Angles in the Triangle**:
- Since \( OAB \) is an equilateral triangle, each angle measures \( 60^\circ \). The angle at the vertex \( O \) is \( 60^\circ \).

4. **Using Trigonometric Relationships**:
- The angle made by the latus rectum with the transverse axis can be considered as \( 30^\circ \) (half of \( 60^\circ \)). Therefore, we can use the tangent function:
\[ \tan(30^\circ) = \frac{b}{aE} \]
- We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \).

5. **Setting Up the Equation**:
- From the tangent relationship, we have:
\[ \frac{b}{aE} = \frac{1}{\sqrt{3}} \]
- Rearranging gives:
\[ b = \frac{aE}{\sqrt{3}} \]

6. **Using the Eccentricity Formula**:
- The relationship between \( a \), \( b \), and the eccentricity \( e \) of the hyperbola is given by:
\[ e^2 = 1 + \frac{b^2}{a^2} \]

7. **Substituting for \( b \)**:
- Substitute \( b \) from the previous equation into the eccentricity formula:
\[ e^2 = 1 + \frac{\left(\frac{aE}{\sqrt{3}}\right)^2}{a^2} \]
- This simplifies to:
\[ e^2 = 1 + \frac{E^2}{3} \]

8. **Rearranging the Equation**:
- Rearranging gives:
\[ 3e^2 = 3 + E^2 \]
- Thus:
\[ E^2 = 3e^2 - 3 \]

9. **Using the Relationship Between Eccentricity and Latus Rectum**:
- From the tangent relationship, we also know:
\[ \tan(30^\circ) = e^2 - 1 \]
- Therefore, substituting gives:
\[ \frac{1}{\sqrt{3}} = e^2 - 1 \]
- Rearranging gives:
\[ e^2 = 1 + \frac{1}{\sqrt{3}} \]

10. **Solving for \( e \)**:
- Now, we can solve for \( e \):
\[ e^2 = \frac{3 + \sqrt{3}}{3} \]
- Taking the square root gives:
\[ e = \sqrt{\frac{3 + \sqrt{3}}{3}} \]

### Final Answer:
The eccentricity \( e \) of the hyperbola is \( \sqrt{\frac{3 + \sqrt{3}}{3}} \).