AB is a diameter of `x^2 + 9y^2=25`. The eccentric angle of A is `pi/6` . Then the eccentric angle of B is

Options

A. `5pi//6`
B. `-5pi//6`
C. `-2pi//3`
D. none of these

Answer

To solve the problem, we need to find the eccentric angle of point B given that point A has an eccentric angle of \( \frac{\pi}{6} \) and that points A and B are the endpoints of a diameter of the ellipse defined by the equation \( x^2 + 9y^2 = 25 \).

### Step-by-Step Solution:

1. **Understanding the Eccentric Angles**:
For any ellipse, if two points A and B are the endpoints of a diameter, the difference in their eccentric angles is \( \pi \) radians. This means that if the eccentric angle of point A is \( \theta_A \), then the eccentric angle of point B, \( \theta_B \), can be expressed as:
\[ \theta_B = \theta_A - \pi \]

2. **Substituting the Given Value**:
We know from the problem that the eccentric angle of point A is:
\[ \theta_A = \frac{\pi}{6} \]
Now, substituting this value into the equation for \( \theta_B \):
\[ \theta_B = \frac{\pi}{6} - \pi \]

3. **Calculating \( \theta_B \)**:
To perform the subtraction, we need to express \( \pi \) in terms of sixths:
\[ \pi = \frac{6\pi}{6} \]
Therefore, we can rewrite \( \theta_B \):
\[ \theta_B = \frac{\pi}{6} - \frac{6\pi}{6} = \frac{\pi - 6\pi}{6} = \frac{-5\pi}{6} \]

4. **Final Result**:
Thus, the eccentric angle of point B is:
\[ \theta_B = -\frac{5\pi}{6} \]

### Conclusion:
The eccentric angle of point B is \( -\frac{5\pi}{6} \).