`P and Q` are points on the ellipse `x^2/a^2+y^2/b^2 =1` whose center is `C`. The eccentric angles of P and Q differ by a right angle. If `/_PCQ` minimum, the eccentric angle of P can be (A) `pi/6` (B) `pi/4` (C) `pi/3` (D) `pi/12`

Options

A. `(pi)/(6)`
B. `(pi)/(4)`
C. `(pi)/(3)`
D. `(pi)/(12)`

Explanation

To solve the problem, we need to find the eccentric angle of point P on the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) given that the eccentric angles of points P and Q differ by \( \frac{\pi}{2} \) radians (90 degrees) and that the angle \( \angle PCQ \) is minimized.

### Step-by-Step Solution:

1. **Define Points P and Q**:
- Let the eccentric angle of point P be \( \theta \).
- The coordinates of point P on the ellipse can be expressed as:
\[ P = (a \cos \theta, b \sin \theta) \]
- Since the eccentric angle of Q differs by \( \frac{\pi}{2} \), we can express the coordinates of point Q as:
\[ Q = (a \cos(\theta + \frac{\pi}{2}), b \sin(\theta + \frac{\pi}{2}) ) \]
- Simplifying this, we get:
\[ Q = (-a \sin \theta, b \cos \theta) \]

2. **Find the Slopes of Lines CP and CQ**:
- The center C of the ellipse is at the origin (0, 0).
- The slope of line CP is given by:
\[ m_{CP} = \frac{y_P}{x_P} = \frac{b \sin \theta}{a \cos \theta} \]
- The slope of line CQ is given by:
\[ m_{CQ} = \frac{y_Q}{x_Q} = \frac{b \cos \theta}{-a \sin \theta} = -\frac{b \cos \theta}{a \sin \theta} \]

3. **Calculate the Angle \( \angle PCQ \)**:
- The angle \( \angle PCQ \) can be calculated using the formula for the tangent of the angle between two lines:
\[ \tan(\angle PCQ) = \left| \frac{m_{CQ} - m_{CP}}{1 + m_{CP} m_{CQ}} \right| \]
- Substituting the slopes:
\[ \tan(\angle PCQ) = \left| \frac{-\frac{b \cos \theta}{a \sin \theta} - \frac{b \sin \theta}{a \cos \theta}}{1 + \left( \frac{b \sin \theta}{a \cos \theta} \cdot -\frac{b \cos \theta}{a \sin \theta} \right)} \right| \]

4. **Minimize the Angle**:
- For \( \angle PCQ \) to be minimized, we need to maximize the sine of the angle, which is related to \( \sin(2\theta) \).
- The maximum value of \( \sin(2\theta) \) occurs when \( 2\theta = \frac{\pi}{2} \) or \( \theta = \frac{\pi}{4} \).

5. **Conclusion**:
- Therefore, the eccentric angle of point P that minimizes \( \angle PCQ \) is:
\[ \theta = \frac{\pi}{4} \]

### Final Answer:
The eccentric angle of P can be \( \frac{\pi}{4} \).