If eccentric angle of a point lying in the first quadrant on the ellipse `x^2/a^2 + y^2/b^2 = 1` be `theta` and the line joining the centre to that point makes an angle `phi` with the x-axis, then `theta - phi` will be maximum when `theta` is equal to

Options

A. `tan^(-1) sqrt((a)/(b))`
B. `tan^(-1)sqrt((b)/(a))`
C. `(pi)/(4)`
D. `(pi)/(3)`

Explanation

To solve the problem, we need to find the value of the eccentric angle \( \theta \) at which the expression \( \theta - \phi \) is maximized, where \( \phi \) is the angle made by the line joining the center of the ellipse to the point \( P \) on the ellipse.

### Step-by-Step Solution:

1. **Identify the Coordinates of Point P**:
The coordinates of point \( P \) on the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) corresponding to the eccentric angle \( \theta \) are:
\[ P(a \cos \theta, b \sin \theta) \]

2. **Determine the Angle \( \phi \)**:
The angle \( \phi \) is the angle that the line joining the center to the point \( P \) makes with the positive x-axis. We can find \( \tan \phi \) using the coordinates of point \( P \):
\[ \tan \phi = \frac{y}{x} = \frac{b \sin \theta}{a \cos \theta} \]
Therefore,
\[ \tan \phi = \frac{b}{a} \tan \theta \]

3. **Express \( \theta - \phi \)**:
We want to maximize \( \theta - \phi \). Using the tangent subtraction formula:
\[ \tan(\theta - \phi) = \frac{\tan \theta - \tan \phi}{1 + \tan \theta \tan \phi} \]
Substituting \( \tan \phi \):
\[ \tan(\theta - \phi) = \frac{\tan \theta - \frac{b}{a} \tan \theta}{1 + \tan \theta \cdot \frac{b}{a} \tan \theta} \]
This simplifies to:
\[ \tan(\theta - \phi) = \frac{\tan \theta \left(1 - \frac{b}{a}\right)}{1 + \frac{b}{a} \tan^2 \theta} \]

4. **Maximizing \( \tan(\theta - \phi) \)**:
To maximize \( \tan(\theta - \phi) \), we need to minimize the denominator:
\[ 1 + \frac{b}{a} \tan^2 \theta \]
This expression will be minimized when \( \tan^2 \theta \) is minimized, which occurs when \( \tan \theta \) is at its critical points.

5. **Setting up the Derivative**:
Differentiate the expression with respect to \( \theta \) and set it to zero:
\[ \frac{d}{d\theta} \left( a \cot \theta + b \tan \theta \right) = 0 \]
This gives:
\[ -a \csc^2 \theta + b \sec^2 \theta = 0 \]
Rearranging gives:
\[ a \csc^2 \theta = b \sec^2 \theta \]

6. **Finding \( \tan \theta \)**:
From the equation \( a \csc^2 \theta = b \sec^2 \theta \), we can derive:
\[ \tan^2 \theta = \frac{a}{b} \]
Therefore,
\[ \tan \theta = \sqrt{\frac{a}{b}} \]

7. **Final Result**:
Thus, the eccentric angle \( \theta \) at which \( \theta - \phi \) is maximized is:
\[ \theta = \tan^{-1} \left( \sqrt{\frac{a}{b}} \right) \]