If the latus rectum of an ellipse is equal to half of minor axis, then its eccentricity is (i) `sqrt((3)/(2))` (ii) `(sqrt(3))/(2)` (iii) `(1)/(2)` (iv) None of these

Options

A. `sqrt((3)/(2))`
B. `(sqrt(3))/(2)`
C. `(1)/(2)`
D. None of these

Answer

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

### Step-by-step Solution:

1. **Understand the properties of the ellipse**:
The standard form of the equation of an ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \( a \) is the semi-major axis and \( b \) is the semi-minor axis.

2. **Identify the latus rectum**:
The length of the latus rectum \( L \) of an ellipse is given by:
\[ L = \frac{2b^2}{a} \]

3. **Set up the relationship**:
According to the problem, the latus rectum is equal to half of the minor axis:
\[ L = \frac{1}{2} \times 2b = b \]
Therefore, we can set up the equation:
\[ \frac{2b^2}{a} = b \]

4. **Simplify the equation**:
Rearranging the equation gives:
\[ 2b^2 = ab \]
Dividing both sides by \( b \) (assuming \( b \neq 0 \)):
\[ 2b = a \]

5. **Use the relationship between \( a \), \( b \), and eccentricity \( e \)**:
We know that:
\[ b^2 = a^2(1 - e^2) \]
Substituting \( a = 2b \) into this equation:
\[ b^2 = (2b)^2(1 - e^2) \]
Simplifying gives:
\[ b^2 = 4b^2(1 - e^2) \]

6. **Rearranging the equation**:
Dividing both sides by \( b^2 \) (assuming \( b \neq 0 \)):
\[ 1 = 4(1 - e^2) \]
Expanding this gives:
\[ 1 = 4 - 4e^2 \]

7. **Solve for \( e^2 \)**:
Rearranging the equation:
\[ 4e^2 = 4 - 1 \]
\[ 4e^2 = 3 \]
Dividing by 4:
\[ e^2 = \frac{3}{4} \]

8. **Finding eccentricity \( e \)**:
Taking the square root of both sides:
\[ e = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

9. **Conclusion**:
The eccentricity of the ellipse is:
\[ e = \frac{\sqrt{3}}{2} \]
Therefore, the correct option is:
**(ii) \(\frac{\sqrt{3}}{2}\)**.