if `2y = x and 3y + 4x=0 ` are the equations of a pair of conjugate diameters of an ellipse , then the eccentricity of the ellipse , is

Options

A. `sqrt(2//3)`
B. `sqrt(2//5)`
C. `sqrt(1//3)`
D. `sqrt(1//2)`

Answer

To find the eccentricity of the ellipse given the equations of a pair of conjugate diameters, we can follow these steps:

### Step 1: Identify the equations of the lines
We have the following equations:
1. \( 2y = x \) (Equation 1)
2. \( 3y + 4x = 0 \) (Equation 2)

### Step 2: Convert the equations to slope-intercept form
For Equation 1:
\[ 2y = x \implies y = \frac{1}{2}x \]
The slope \( m_1 = \frac{1}{2} \).

For Equation 2:
\[ 3y + 4x = 0 \implies 3y = -4x \implies y = -\frac{4}{3}x \]
The slope \( m_2 = -\frac{4}{3} \).

### Step 3: Use the relationship for conjugate diameters
For the ellipse, the relationship between the slopes of the conjugate diameters is given by:
\[ m_1 \cdot m_2 = -\frac{b^2}{a^2} \]
Substituting the values of \( m_1 \) and \( m_2 \):
\[ \frac{1}{2} \cdot -\frac{4}{3} = -\frac{b^2}{a^2} \]
Calculating the left side:
\[ -\frac{4}{6} = -\frac{2}{3} \]
Thus, we have:
\[ \frac{b^2}{a^2} = \frac{2}{3} \]

### Step 4: Find the eccentricity of the ellipse
The eccentricity \( e \) of the ellipse is given by the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting \( \frac{b^2}{a^2} = \frac{2}{3} \):
\[ e = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} \]

### Conclusion
Thus, the eccentricity of the ellipse is:
\[ e = \frac{1}{\sqrt{3}} \]