The ratio of the ordinates of a point and its .corresponding point is`(2sqrt(2))/(3)` then eccentricity is

Options

A. `1/3`
B. `2/3`
C. `sqrt(2)/(3)`
D. `(2sqrt(2))/(3)`

Answer

To solve the problem, we need to find the eccentricity of an ellipse given the ratio of the ordinates of a point and its corresponding point on the ellipse.

### Step-by-Step Solution:

1. **Understand the Points on the Ellipse**:
The points on the ellipse can be represented in parametric form as:
- Point P: \( (A \cos \theta, B \sin \theta) \)
- Corresponding Point P1: \( (A \sin \theta, B \cos \theta) \)

2. **Set Up the Ratio of Ordinates**:
The ordinates (y-coordinates) of the points P and P1 are \( B \sin \theta \) and \( B \cos \theta \), respectively. According to the problem, the ratio of these ordinates is given as:
\[ \frac{B \sin \theta}{B \cos \theta} = \frac{2\sqrt{2}}{3} \]

3. **Simplify the Ratio**:
Since \( B \) is common in both the numerator and the denominator, we can cancel it out:
\[ \frac{\sin \theta}{\cos \theta} = \frac{2\sqrt{2}}{3} \]
This simplifies to:
\[ \tan \theta = \frac{2\sqrt{2}}{3} \]

4. **Find the Ratio of Semi-Axes**:
From the ratio of the ordinates, we can express the ratio of the semi-minor axis \( B \) to the semi-major axis \( A \):
\[ \frac{B}{A} = \frac{2\sqrt{2}}{3} \]

5. **Square the Ratio**:
Squaring both sides gives:
\[ \frac{B^2}{A^2} = \left(\frac{2\sqrt{2}}{3}\right)^2 = \frac{8}{9} \]

6. **Use the Eccentricity Formula**:
The formula for the eccentricity \( e \) of an ellipse is given by:
\[ e = \sqrt{1 - \frac{B^2}{A^2}} \]
Substituting the value we found:
\[ e = \sqrt{1 - \frac{8}{9}} = \sqrt{\frac{1}{9}} = \frac{1}{3} \]

7. **Final Answer**:
Therefore, the eccentricity of the ellipse is:
\[ e = \frac{1}{3} \]