if `theta and phi` are eccentric angles of the ends of a pair of conjugate diameters of the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` then `( theta - phi)` is equal to

Options

A. `+- pi //2`
B. `+- pi`
C. 0
D. none of these

Answer

To solve the problem, we need to find the value of \( \theta - \phi \) where \( \theta \) and \( \phi \) are the eccentric angles of the ends of a pair of conjugate diameters of the ellipse given by the equation:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

### Step 1: Understanding the Eccentric Angles
The eccentric angles \( \theta \) and \( \phi \) correspond to points on the ellipse. For any point on the ellipse, the coordinates can be expressed in terms of the eccentric angle \( \theta \) as:
\[ x = a \cos \theta, \quad y = b \sin \theta \]

### Step 2: Equations of the Conjugate Diameters
The equations of the conjugate diameters can be expressed using the tangent equations corresponding to \( \theta \) and \( \phi \):
1. For angle \( \theta \):
\[ \frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1 \]
2. For angle \( \phi \):
\[ \frac{x}{a} \cos \phi + \frac{y}{b} \sin \phi = 1 \]

### Step 3: Finding the Slopes of the Diameters
From the tangent equations, we can derive the slopes of the lines corresponding to \( \theta \) and \( \phi \):
- The slope \( m_1 \) corresponding to \( \theta \):
\[ m_1 = -\frac{b \cos \theta}{a \sin \theta} \]
- The slope \( m_2 \) corresponding to \( \phi \):
\[ m_2 = -\frac{b \cos \phi}{a \sin \phi} \]

### Step 4: Condition for Conjugate Diameters
For conjugate diameters, the product of their slopes should satisfy:
\[ m_1 \cdot m_2 = -\frac{b^2}{a^2} \]

Substituting the expressions for \( m_1 \) and \( m_2 \):
\[ \left(-\frac{b \cos \theta}{a \sin \theta}\right) \left(-\frac{b \cos \phi}{a \sin \phi}\right) = -\frac{b^2}{a^2} \]
This simplifies to:
\[ \frac{b^2 \cos \theta \cos \phi}{a^2 \sin \theta \sin \phi} = -\frac{b^2}{a^2} \]

### Step 5: Simplifying the Equation
Cancelling \( \frac{b^2}{a^2} \) from both sides gives:
\[ \frac{\cos \theta \cos \phi}{\sin \theta \sin \phi} = -1 \]
This can be rewritten as:
\[ \cot \theta \cot \phi = -1 \]

### Step 6: Solving for \( \theta - \phi \)
From the equation \( \cot \theta \cot \phi = -1 \), we can conclude:
\[ \cot \theta = -\cot \phi \]
This implies:
\[ \theta = 90^\circ + \phi \quad \text{or} \quad \theta = \frac{\pi}{2} + \phi \quad (\text{in radians}) \]

### Step 7: Final Result
Thus, we find:
\[ \theta - \phi = \frac{\pi}{2} \]

### Conclusion
The value of \( \theta - \phi \) is:
\[ \theta - \phi = \frac{\pi}{2} \]