If the latus rectum of a hyperbola subtend an angle of `60A^(@)` at the other focus, then eccentricity of the hyperbola is

Options

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

Answer

To solve the problem, we need to find the eccentricity of a hyperbola given that the latus rectum subtends an angle of \(60^\circ\) at the other focus.

### Step-by-Step Solution:

1. **Understand the Geometry**: The latus rectum of the hyperbola subtends an angle of \(60^\circ\) at the other focus. This means that the angle between the lines drawn from the other focus to the endpoints of the latus rectum is \(60^\circ\).

2. **Define the Hyperbola**: The standard form of the hyperbola is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
where \(a\) is the semi-major axis and \(b\) is the semi-minor axis.

3. **Latus Rectum Length**: The length of the latus rectum \(L\) of a hyperbola is given by:
\[ L = \frac{2b^2}{a} \]

4. **Angle and Slopes**: The angle subtended by the latus rectum at the focus can be related to the slopes of the lines from the focus to the endpoints of the latus rectum. Since the angle is \(60^\circ\), the half-angle is \(30^\circ\).

5. **Slope Calculation**: The slope \(m\) of the lines from the focus to the endpoints of the latus rectum can be expressed as:
\[ m = \tan(30^\circ) = \frac{1}{\sqrt{3}} \]

6. **Relate Slope to Latus Rectum**: The slope can also be expressed in terms of \(b\) and \(e\) (eccentricity):
\[ m = \frac{b^2}{2ae} \]
Setting the two expressions for \(m\) equal gives:
\[ \frac{b^2}{2ae} = \frac{1}{\sqrt{3}} \]

7. **Rearranging**: Rearranging the equation gives:
\[ b^2 = \frac{2ae}{\sqrt{3}} \]

8. **Using the Relationship Between \(b\) and \(e\)**: We know that:
\[ b^2 = a^2(e^2 - 1) \]
Setting the two expressions for \(b^2\) equal gives:
\[ a^2(e^2 - 1) = \frac{2ae}{\sqrt{3}} \]

9. **Eliminate \(a\)**: Dividing both sides by \(a\) (assuming \(a \neq 0\)):
\[ a(e^2 - 1) = \frac{2e}{\sqrt{3}} \]

10. **Rearranging to Form a Quadratic Equation**: Rearranging gives:
\[ e^2 - 1 - \frac{2e}{a\sqrt{3}} = 0 \]
Let \(k = \frac{2}{\sqrt{3}}\), then:
\[ e^2 - ke - 1 = 0 \]

11. **Using the Quadratic Formula**: The roots of the quadratic equation \(e^2 - ke - 1 = 0\) can be found using the quadratic formula:
\[ e = \frac{k \pm \sqrt{k^2 + 4}}{2} \]
Substituting \(k = \frac{2}{\sqrt{3}}\):
\[ e = \frac{\frac{2}{\sqrt{3}} \pm \sqrt{\left(\frac{2}{\sqrt{3}}\right)^2 + 4}}{2} \]

12. **Simplifying**: This simplifies to:
\[ e = \frac{\frac{2}{\sqrt{3}} \pm \sqrt{\frac{4}{3} + 4}}{2} = \frac{\frac{2}{\sqrt{3}} \pm \sqrt{\frac{4 + 12}{3}}}{2} = \frac{\frac{2}{\sqrt{3}} \pm \sqrt{\frac{16}{3}}}{2} \]
\[ = \frac{\frac{2}{\sqrt{3}} \pm \frac{4}{\sqrt{3}}}{2} = \frac{6/\sqrt{3}}{2} = \frac{3}{\sqrt{3}} = \sqrt{3} \]

### Final Answer:
Thus, the eccentricity \(e\) of the hyperbola is:
\[ e = \sqrt{3} \]