The eccentricity of an ellipse whose pair of conjugate diameters are `2y = x `and `3y = -2x `is

Options

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

Answer

To find the eccentricity of the ellipse whose pair of conjugate diameters are given by the equations \(2y = x\) and \(3y = -2x\), we will follow these steps:

### Step 1: Identify the slopes of the diameters
The equations of the diameters can be rewritten in slope-intercept form (i.e., \(y = mx\)):
1. From \(2y = x\), we get:
\[ y = \frac{1}{2}x \quad \text{(slope } m_1 = \frac{1}{2}\text{)} \]
2. From \(3y = -2x\), we get:
\[ y = -\frac{2}{3}x \quad \text{(slope } m_2 = -\frac{2}{3}\text{)} \]

### Step 2: Use the relationship between slopes and the ellipse parameters
For conjugate diameters, the product of their slopes is given by:
\[ m_1 \cdot m_2 = -\frac{B^2}{A^2} \]
Substituting the values of \(m_1\) and \(m_2\):
\[ \frac{1}{2} \cdot \left(-\frac{2}{3}\right) = -\frac{B^2}{A^2} \]
Calculating the left side:
\[ -\frac{1}{3} = -\frac{B^2}{A^2} \]
Thus, we have:
\[ \frac{B^2}{A^2} = \frac{1}{3} \]

### Step 3: Calculate the eccentricity
The eccentricity \(e\) of an ellipse is given by the formula:
\[ e = \sqrt{1 - \frac{B^2}{A^2}} \]
Substituting the value we found:
\[ e = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}} \]

### Final Answer
Thus, the eccentricity of the ellipse is:
\[ e = \sqrt{\frac{2}{3}} \]