The eccentric angle of the point where the line, `5x - 3y = 8sqrt2` is a normal to the ellipse `x^2/25 + y^2/9 =1` is

Options

A. `(3pi)/4`
B. `pi/4`
C. `pi/6`
D. `tan^(-1)(1/2)`

Answer

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

### Step-by-Step Solution:

1. **Identify 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:
- \(A = 5\) (since \(A^2 = 25\))
- \(B = 3\) (since \(B^2 = 9\))

2. **Find the Normal Equation:**
The normal to the ellipse at the point \((x_1, y_1)\) can be expressed as:
\[ \frac{A^2 x}{x_1} - \frac{B^2 y}{y_1} = A^2 - B^2 \]
Substituting \(A^2 = 25\) and \(B^2 = 9\):
\[ \frac{25x}{x_1} - \frac{9y}{y_1} = 16 \]

3. **Parameterize the Point on the Ellipse:**
The point on the ellipse can be expressed in terms of the eccentric angle \(\theta\):
\[ x_1 = A \cos \theta = 5 \cos \theta \]
\[ y_1 = B \sin \theta = 3 \sin \theta \]

4. **Substitute \(x_1\) and \(y_1\) into the Normal Equation:**
Substitute \(x_1\) and \(y_1\) into the normal equation:
\[ \frac{25x}{5 \cos \theta} - \frac{9y}{3 \sin \theta} = 16 \]
Simplifying this gives:
\[ 5 \frac{x}{\cos \theta} - 3 \frac{y}{\sin \theta} = 16 \]

5. **Compare with the Given Line Equation:**
The given line equation is:
\[ 5x - 3y = 8\sqrt{2} \]
We can rewrite this as:
\[ 5x - 3y - 8\sqrt{2} = 0 \]
Now, we can compare the coefficients of \(x\) and \(y\) from both equations:
\[ 5 \frac{1}{\cos \theta} = 5 \quad \text{and} \quad -3 \frac{1}{\sin \theta} = -3 \]
This leads to:
\[ \frac{1}{\cos \theta} = 1 \quad \text{and} \quad \frac{1}{\sin \theta} = 1 \]
From the constant terms:
\[ 16 = 8\sqrt{2} \]

6. **Solve for \(\cos \theta\) and \(\sin \theta\):**
From the equations:
\[ \cos \theta = \frac{1}{\sqrt{2}} \quad \text{and} \quad \sin \theta = \frac{1}{\sqrt{2}} \]

7. **Determine \(\theta\):**
Since \(\cos \theta = \sin \theta = \frac{1}{\sqrt{2}}\), we can conclude:
\[ \theta = \frac{\pi}{4} \]

### Final Answer:
The eccentric angle of the point where the line \(5x - 3y = 8\sqrt{2}\) is a normal to the ellipse is:
\[ \theta = \frac{\pi}{4} \]