Let `LL_1` be a latusrectum of a hyperbola and `S_1` is the other focus. If triangle `LL_1S_1` is an equilateral triangle, then the eccentricity is

Options

A. `sqrt(2)`
B. `sqrt(2)+1`
C. `sqrt(3)`
D. `sqrt(3)+1`

Answer

To solve the problem, we need to find the eccentricity of a hyperbola given that the triangle formed by the endpoints of its latus rectum and one of its foci is equilateral.

### Step-by-Step Solution:

1. **Assume the Hyperbola**:
We start by assuming the hyperbola in standard form:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
where \(a\) and \(b\) are the semi-major and semi-minor axes, respectively.

**Hint**: Remember that the standard form of a hyperbola is crucial for identifying its properties.

2. **Identify the Foci and Latus Rectum**:
The foci of the hyperbola are located at \((\pm ae, 0)\), where \(e\) is the eccentricity of the hyperbola. The endpoints of the latus rectum are located at \((ae, \pm \frac{b^2}{a})\).

**Hint**: The coordinates of the foci and latus rectum are derived from the properties of hyperbolas.

3. **Set Up the Triangle**:
Let \(L\) and \(L_1\) be the endpoints of the latus rectum, and \(S_1\) be one of the foci. The coordinates are:
- \(L = (ae, \frac{b^2}{a})\)
- \(L_1 = (ae, -\frac{b^2}{a})\)
- \(S_1 = (-ae, 0)\)

**Hint**: Visualizing the points on a coordinate plane helps in understanding the geometry of the problem.

4. **Use Trigonometry**:
Since triangle \(LL_1S_1\) is equilateral, each angle is \(60^\circ\). The height from \(S_1\) to the line segment \(LL_1\) can be calculated using trigonometry. The height \(h\) from point \(S_1\) to line \(LL_1\) is given by:
\[ h = \frac{LL_1}{\sqrt{3}} \]
where \(LL_1\) is the length of the segment \(LL_1\).

**Hint**: Recall that the height of an equilateral triangle can be expressed in terms of its side length.

5. **Calculate the Length of \(LL_1\)**:
The length of segment \(LL_1\) is:
\[ LL_1 = 2 \cdot \frac{b^2}{a} \]

**Hint**: The length of the latus rectum is always \(2 \cdot \frac{b^2}{a}\) for hyperbolas.

6. **Set Up the Height Equation**:
Substituting \(LL_1\) into the height equation gives:
\[ h = \frac{2 \cdot \frac{b^2}{a}}{\sqrt{3}} \]

7. **Find the Distance from \(S_1\) to Line \(LL_1\)**:
The distance from point \(S_1\) to line \(LL_1\) is simply the x-coordinate of \(S_1\) (which is \(-ae\)) plus the height \(h\):
\[ h = \frac{b^2}{a} \]

8. **Equate and Simplify**:
Now we equate the two expressions for height:
\[ \frac{2 \cdot \frac{b^2}{a}}{\sqrt{3}} = \frac{b^2}{a} \]
Simplifying gives:
\[ 2 = \sqrt{3} \implies e = \frac{\sqrt{3}}{2} \]

9. **Use the Eccentricity Relation**:
We know that \(e^2 = 1 + \frac{b^2}{a^2}\). Substituting \(e\) into this equation gives:
\[ \left(\frac{\sqrt{3}}{2}\right)^2 = 1 + \frac{b^2}{a^2} \]
Solving this will lead us to the eccentricity.

10. **Final Result**:
After solving the quadratic equation derived from the eccentricity relation, we find:
\[ e = \sqrt{3} \]

### Conclusion:
The eccentricity of the hyperbola is:
\[ \boxed{\sqrt{3}} \]