The ecentric angle of the points on the ellipse `(x^(2))/(6)+(y^(2))/(2)=1` whose distane from its centre is 2 is/are.

Options

A. `(pi)/(4)`
B. `(3pi)/(4)`
C. `(5pi)/(4)`
D. `(7pi)/(4)`

Answer

To solve the problem, we need to find the eccentric angles of the points on the ellipse \(\frac{x^2}{6} + \frac{y^2}{2} = 1\) that are at a distance of 2 from the center of the ellipse.

### Step-by-Step Solution:

1. **Identify the Standard Form of the Ellipse**:
The given equation of the ellipse is \(\frac{x^2}{6} + \frac{y^2}{2} = 1\). This 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, \(a = \sqrt{6}\) and \(b = \sqrt{2}\).

2. **Parametric Equations of the Ellipse**:
The parametric equations for points on the ellipse can be expressed using the eccentric angle \(\theta\):
\[ x = a \cos \theta = \sqrt{6} \cos \theta \]
\[ y = b \sin \theta = \sqrt{2} \sin \theta \]

3. **Distance from the Center**:
The distance \(OP\) from the center (0,0) to a point \(P(x,y)\) on the ellipse is given by:
\[ OP = \sqrt{x^2 + y^2} \]
Substituting the parametric equations:
\[ OP = \sqrt{(\sqrt{6} \cos \theta)^2 + (\sqrt{2} \sin \theta)^2} \]
Simplifying this:
\[ OP = \sqrt{6 \cos^2 \theta + 2 \sin^2 \theta} \]

4. **Set the Distance Equal to 2**:
We know the distance is 2, so we set up the equation:
\[ \sqrt{6 \cos^2 \theta + 2 \sin^2 \theta} = 2 \]
Squaring both sides:
\[ 6 \cos^2 \theta + 2 \sin^2 \theta = 4 \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ 6 \cos^2 \theta + 2 (1 - \cos^2 \theta) = 4 \]
Simplifying further:
\[ 6 \cos^2 \theta + 2 - 2 \cos^2 \theta = 4 \]
\[ 4 \cos^2 \theta = 2 \]
\[ \cos^2 \theta = \frac{1}{2} \]

6. **Finding the Values of \(\theta\)**:
Taking the square root:
\[ \cos \theta = \frac{1}{\sqrt{2}} \quad \text{or} \quad \cos \theta = -\frac{1}{\sqrt{2}} \]
This gives:
\[ \theta = \frac{\pi}{4}, \quad \frac{7\pi}{4} \quad (\text{for } \cos \theta = \frac{1}{\sqrt{2}}) \]
\[ \theta = \frac{3\pi}{4}, \quad \frac{5\pi}{4} \quad (\text{for } \cos \theta = -\frac{1}{\sqrt{2}}) \]

7. **Final Eccentric Angles**:
Thus, the eccentric angles of the points on the ellipse whose distance from the center is 2 are:
\[ \theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \]