If the extremities of a focal chord are `(5pi)/(12)` and `(pi)/(12)` then e =

Options

A. `(1)/sqrt(2)`
B. `sqrt(3/2)`
C. `sqrt(3)/(4)`
D. `sqrt(3)/(2)`

Answer

To find the eccentricity \( e \) of the ellipse given the extremities of a focal chord at \( \frac{5\pi}{12} \) and \( \frac{\pi}{12} \), we can use the formula related to focal chords:

\[ E \cdot \cos\left(\frac{\alpha + \beta}{2}\right) = \cos\left(\frac{\alpha - \beta}{2}\right) \]

### Step-by-Step Solution:

1. **Identify the values of \( \alpha \) and \( \beta \)**:
- Given \( \alpha = \frac{5\pi}{12} \) and \( \beta = \frac{\pi}{12} \).

2. **Calculate \( \alpha + \beta \)**:
\[ \alpha + \beta = \frac{5\pi}{12} + \frac{\pi}{12} = \frac{6\pi}{12} = \frac{\pi}{2} \]

3. **Calculate \( \alpha - \beta \)**:
\[ \alpha - \beta = \frac{5\pi}{12} - \frac{\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3} \]

4. **Substitute into the formula**:
\[ E \cdot \cos\left(\frac{\pi}{2} \cdot \frac{1}{2}\right) = \cos\left(\frac{\pi}{3} \cdot \frac{1}{2}\right) \]
This simplifies to:
\[ E \cdot \cos\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{6}\right) \]

5. **Evaluate \( \cos\left(\frac{\pi}{4}\right) \) and \( \cos\left(\frac{\pi}{6}\right) \)**:
- \( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \)
- \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \)

6. **Substitute these values back into the equation**:
\[ E \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{3}}{2} \]

7. **Solve for \( E \)**:
\[ E = \frac{\sqrt{3}}{2} \cdot \sqrt{2} = \frac{\sqrt{6}}{2} \]

8. **Final Result**:
\[ E = \frac{\sqrt{6}}{2} \]