The line `5x-3y=8sqrt2` is a normal to the ellipse `x^(2)/25+y^(2)/9=1`, If 'theta' be eccentric angle of the foot of this normal then 'theta' is equal to

Options

A. `pi/6`
B. `pi/4`
C. `pi/3`
D. `pi/2`

Answer

To solve the problem, we need to find the eccentric angle \( \theta \) of the foot of the normal to the given ellipse at the point where the line \( 5x - 3y = 8\sqrt{2} \) is normal to the ellipse \( \frac{x^2}{25} + \frac{y^2}{9} = 1 \).

### Step-by-Step Solution:

1. **Identify the parameters of the ellipse:**
The equation of the ellipse is given as:
\[ \frac{x^2}{25} + \frac{y^2}{9} = 1 \]
From this, we can identify:
- \( a^2 = 25 \) → \( a = 5 \)
- \( b^2 = 9 \) → \( b = 3 \)

2. **Write the parametric equations of the ellipse:**
A point on the ellipse can be represented parametrically as:
\[ x = a \cos \theta = 5 \cos \theta \]
\[ y = b \sin \theta = 3 \sin \theta \]

3. **Write the equation of the normal at the point \( (5 \cos \theta, 3 \sin \theta) \):**
The equation of the normal to the ellipse at this point is given by:
\[ \frac{x}{a} \cdot \sec \theta - \frac{y}{b} \cdot \csc \theta = a^2 - b^2 \]
Substituting \( a \) and \( b \):
\[ 5x \sec \theta - 3y \csc \theta = 16 \]

4. **Set up the equation of the normal:**
This can be rearranged as:
\[ 5x \sec \theta - 3y \csc \theta - 16 = 0 \]
This is our equation of the normal (let's call it Equation 1).

5. **Compare with the given normal line:**
The given normal line is:
\[ 5x - 3y = 8\sqrt{2} \]
Rearranging gives:
\[ 5x - 3y - 8\sqrt{2} = 0 \]
This is our Equation 2.

6. **Equate coefficients from both equations:**
From Equations 1 and 2, we can compare coefficients:
\[ 5 \sec \theta = 8\sqrt{2} \quad \text{and} \quad 3 \csc \theta = 8\sqrt{2} \]

7. **Solve for \( \sec \theta \) and \( \csc \theta \):**
From \( 5 \sec \theta = 8\sqrt{2} \):
\[ \sec \theta = \frac{8\sqrt{2}}{5} \]
From \( 3 \csc \theta = 8\sqrt{2} \):
\[ \csc \theta = \frac{8\sqrt{2}}{3} \]

8. **Use the identity \( \sec^2 \theta - 1 = \tan^2 \theta \) and \( \csc^2 \theta - 1 = \cot^2 \theta \):**
We know:
\[ \sec^2 \theta = 1 + \tan^2 \theta \quad \text{and} \quad \csc^2 \theta = 1 + \cot^2 \theta \]
From \( \sec^2 \theta \) and \( \csc^2 \theta \), we can find \( \sin \theta \) and \( \cos \theta \).

9. **Find \( \sin \theta \) and \( \cos \theta \):**
Since \( \sec \theta = \frac{1}{\cos \theta} \) and \( \csc \theta = \frac{1}{\sin \theta} \):
\[ \sin \theta = \frac{1}{\csc \theta} = \frac{3}{8\sqrt{2}} \quad \text{and} \quad \cos \theta = \frac{1}{\sec \theta} = \frac{5}{8\sqrt{2}} \]

10. **Determine \( \theta \):**
Since \( \sin \theta = \cos \theta \), we find:
\[ \theta = 45^\circ = \frac{\pi}{4} \]

### Final Answer:
Thus, the eccentric angle \( \theta \) is:
\[ \theta = \frac{\pi}{4} \]