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 triangle \( LL_1S_1 \) is an equilateral triangle, where \( LL_1 \) is a latus rectum of the hyperbola and \( S_1 \) is the other focus.

### Step-by-Step Solution:

1. **Understanding the Hyperbola**:
The standard equation of a hyperbola is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
where \( 2c \) is the distance between the foci, \( c^2 = a^2 + b^2 \), and the eccentricity \( e \) is defined as \( e = \frac{c}{a} \).

2. **Coordinates of the Focus and Latus Rectum**:
The foci of the hyperbola are located at \( (ae, 0) \) and \( (-ae, 0) \). The endpoints of the latus rectum \( L \) and \( L_1 \) can be found at:
\[ L = \left( ae, b\sqrt{e^2 - 1} \right) \quad \text{and} \quad L_1 = \left( ae, -b\sqrt{e^2 - 1} \right) \]

3. **Setting Up the Triangle**:
The points of the triangle are \( L, L_1, \) and \( S_1 \) where \( S_1 \) is the other focus at \( (-ae, 0) \). Thus, the coordinates are:
- \( L = \left( ae, b\sqrt{e^2 - 1} \right) \)
- \( L_1 = \left( ae, -b\sqrt{e^2 - 1} \right) \)
- \( S_1 = (-ae, 0) \)

4. **Finding Lengths of the Sides**:
- Length \( LL_1 \):
\[ LL_1 = 2b\sqrt{e^2 - 1} \]
- Length \( LS_1 \):
\[ LS_1 = \sqrt{(ae - (-ae))^2 + \left(b\sqrt{e^2 - 1} - 0\right)^2} = \sqrt{(2ae)^2 + (b\sqrt{e^2 - 1})^2} \]
- Length \( L_1S_1 \):
\[ L_1S_1 = \sqrt{(ae - (-ae))^2 + \left(-b\sqrt{e^2 - 1} - 0\right)^2} = \sqrt{(2ae)^2 + (b\sqrt{e^2 - 1})^2} \]

5. **Equilateral Triangle Condition**:
Since triangle \( LL_1S_1 \) is equilateral, we have:
\[ LL_1 = LS_1 \]
Thus,
\[ 2b\sqrt{e^2 - 1} = \sqrt{(2ae)^2 + (b\sqrt{e^2 - 1})^2} \]

6. **Squaring Both Sides**:
Squaring both sides gives:
\[ 4b^2(e^2 - 1) = 4a^2e^2 + b^2(e^2 - 1) \]
Rearranging this yields:
\[ 4b^2e^2 - 4b^2 = 4a^2e^2 + b^2e^2 - b^2 \]
Simplifying further:
\[ (4b^2 - b^2)e^2 = 4a^2e^2 + 4b^2 \]
\[ 3b^2e^2 = 4a^2e^2 + 4b^2 \]

7. **Substituting \( b^2 \)**:
Using the relation \( b^2 = a^2(e^2 - 1) \):
\[ 3a^2(e^2 - 1)e^2 = 4a^2e^2 + 4a^2(e^2 - 1) \]
Dividing through by \( a^2 \) (assuming \( a \neq 0 \)):
\[ 3(e^2 - 1)e^2 = 4e^2 + 4(e^2 - 1) \]
Simplifying gives:
\[ 3e^4 - 3e^2 = 4e^2 + 4e^2 - 4 \]
\[ 3e^4 - 11e^2 + 4 = 0 \]

8. **Solving the Quadratic**:
Using the quadratic formula:
\[ e^2 = \frac{11 \pm \sqrt{121 - 48}}{6} = \frac{11 \pm \sqrt{73}}{6} \]
Since \( e \) must be positive, we take the positive root.

9. **Final Result**:
The eccentricity \( e \) is:
\[ e = \sqrt{\frac{11 + \sqrt{73}}{6}} \]