If `x/a + y/b = sqrt(2)` touches the ellipse `(x^2)/(a^2) + (y^2)/(b^2) = 1` at a point P, then eccentric angle of P is

Options

A. `0`
B. `45^@`
C. `60^@`
D. `90^@`

Answer

To solve the problem, we need to find the eccentric angle of the point \( P \) where the line \( \frac{x}{a} + \frac{y}{b} = \sqrt{2} \) touches the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).

### Step-by-step Solution:

1. **Identify the given equations**:
- The equation of the ellipse is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
- The equation of the line is:
\[ \frac{x}{a} + \frac{y}{b} = \sqrt{2} \]

2. **Rewrite the equation of the line**:
- We can rewrite the line equation in the standard form:
\[ y = -\frac{b}{a}x + b\sqrt{2} \]

3. **Find the slope of the tangent line**:
- The slope of the tangent line is \( -\frac{b}{a} \).

4. **Use the tangent equation of the ellipse**:
- The equation of the tangent to the ellipse at the point \( P \) with eccentric angle \( \theta \) is given by:
\[ \frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1 \]

5. **Equate the two equations**:
- From the tangent equation, we have:
\[ \frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1 \]
- Substitute \( \frac{x}{a} + \frac{y}{b} = \sqrt{2} \) into the tangent equation:
\[ \cos \theta + \sin \theta = \sqrt{2} \]

6. **Solve for \( \cos \theta \) and \( \sin \theta \)**:
- We can set:
\[ \cos \theta = \frac{1}{\sqrt{2}}, \quad \sin \theta = \frac{1}{\sqrt{2}} \]

7. **Determine the eccentric angle**:
- Since \( \cos \theta = \sin \theta = \frac{1}{\sqrt{2}} \), we find that:
\[ \theta = 45^\circ \]

### Final Answer:
The eccentric angle of point \( P \) is \( 45^\circ \).