If the latus rectum of an ellipse is equal to half of minor axis, then its eccentricity is

Options

A. (a) `(3)/(2)`
B. (b) `(1)/( sqrt(2))`
C. (c) ` (sqrt(3))/( 2)`
D. (d) `(1)/(2)`

Answer

To solve the problem, we need to find the eccentricity of an ellipse given that the latus rectum is equal to half of the minor axis.

### Step-by-Step Solution:

1. **Understanding the Definitions**:
- The equation of an ellipse in standard form is given by:
\[ \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1 \]
- The length of the latus rectum (L) of the ellipse is given by:
\[ L = \frac{2B^2}{A} \]
- The length of the minor axis is given by:
\[ \text{Minor Axis} = 2B \]

2. **Setting Up the Equation**:
- According to the problem, the latus rectum is equal to half of the minor axis:
\[ \frac{2B^2}{A} = \frac{1}{2} \times 2B \]
- Simplifying the right side gives:
\[ \frac{2B^2}{A} = B \]

3. **Eliminating Common Terms**:
- We can multiply both sides by \( A \) (assuming \( A \neq 0 \)):
\[ 2B^2 = AB \]
- Dividing both sides by \( B \) (assuming \( B \neq 0 \)):
\[ 2B = A \]

4. **Finding the Eccentricity**:
- The formula for the eccentricity (E) of the ellipse is:
\[ E = \sqrt{1 - \frac{B^2}{A^2}} \]
- Substituting \( A = 2B \) into the eccentricity formula:
\[ E = \sqrt{1 - \frac{B^2}{(2B)^2}} = \sqrt{1 - \frac{B^2}{4B^2}} \]
- This simplifies to:
\[ E = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

5. **Conclusion**:
- The eccentricity of the ellipse is:
\[ E = \frac{\sqrt{3}}{2} \]
- Therefore, the correct answer is option C: \( \frac{\sqrt{3}}{2} \).