The eccentric angle of a point on the ellipse `x^(2)/6 + y^(2)/2 = 1` whose distance from the centre of the ellipse is 2, is

Options

A. `pi//4`
B. `3pi//2`
C. `5pi//3`
D. `7pi//6`

Explanation

To find the eccentric angle of a point on the ellipse given by the equation \( \frac{x^2}{6} + \frac{y^2}{2} = 1 \) whose distance from the center of the ellipse is 2, we can follow these steps:

### Step 1: Identify the semi-major and semi-minor axes
The given equation of the ellipse can be rewritten in the standard form:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \( a^2 = 6 \) and \( b^2 = 2 \). Thus, we have:
\[ a = \sqrt{6}, \quad b = \sqrt{2} \]

### Step 2: Parametrize the ellipse
A point on the ellipse can be represented using the eccentric angle \( \theta \):
\[ x = a \cos \theta = \sqrt{6} \cos \theta, \quad y = b \sin \theta = \sqrt{2} \sin \theta \]

### Step 3: Calculate the distance from the center
The distance \( d \) from the center of the ellipse to the point \( (x, y) \) is given by:
\[ d = \sqrt{x^2 + y^2} \]
Substituting the parametric equations:
\[ d = \sqrt{(\sqrt{6} \cos \theta)^2 + (\sqrt{2} \sin \theta)^2} \]
This simplifies to:
\[ d = \sqrt{6 \cos^2 \theta + 2 \sin^2 \theta} \]

### Step 4: Set the distance equal to 2
According to the problem, this distance is given as 2:
\[ \sqrt{6 \cos^2 \theta + 2 \sin^2 \theta} = 2 \]
Squaring both sides, we get:
\[ 6 \cos^2 \theta + 2 \sin^2 \theta = 4 \]

### Step 5: Rearrange the equation
Rearranging the equation gives:
\[ 6 \cos^2 \theta + 2 \sin^2 \theta - 4 = 0 \]
We can express \( \sin^2 \theta \) in terms of \( \cos^2 \theta \) using \( \sin^2 \theta = 1 - \cos^2 \theta \):
\[ 6 \cos^2 \theta + 2(1 - \cos^2 \theta) - 4 = 0 \]
This simplifies to:
\[ 6 \cos^2 \theta + 2 - 2 \cos^2 \theta - 4 = 0 \]
\[ 4 \cos^2 \theta - 2 = 0 \]
\[ 4 \cos^2 \theta = 2 \]
\[ \cos^2 \theta = \frac{1}{2} \]

### Step 6: Solve for \( \theta \)
Taking the square root gives:
\[ \cos \theta = \pm \frac{1}{\sqrt{2}} \quad \Rightarrow \quad \theta = \frac{\pi}{4}, \frac{3\pi}{4} \text{ (and their respective angles in other quadrants)} \]

### Conclusion
Thus, the eccentric angles are:
\[ \theta = \frac{\pi}{4} \text{ and } \theta = \frac{3\pi}{4} \]