The latus rectum of an ellipse is half of its minor axis. Its eccentricity is :

Options

A. `(1)/(4)`
B. `(sqrt(3))/(4)`
C. `(1)/(2)`
D. `(sqrt(3))/(2)`

Answer

To find the eccentricity of the ellipse given that the latus rectum is half of its minor axis, we can follow these steps:

### Step 1: Define the ellipse
We start with the standard equation of the ellipse:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \( a > b \).

### Step 2: Write the formula for the latus rectum
The length of the latus rectum \( L \) of an ellipse is given by:
\[ L = \frac{2b^2}{a} \]

### Step 3: Determine the length of the minor axis
The length of the minor axis of the ellipse is:
\[ \text{Minor Axis} = 2b \]

### Step 4: Set up the equation based on the problem statement
According to the problem, the latus rectum is half of the minor axis:
\[ \frac{2b^2}{a} = \frac{1}{2} \times 2b \]
This simplifies to:
\[ \frac{2b^2}{a} = b \]

### Step 5: Simplify the equation
We can simplify the equation by multiplying both sides by \( a \) and dividing by \( b \) (assuming \( b \neq 0 \)):
\[ 2b^2 = ab \]
Dividing both sides by \( b \):
\[ 2b = a \]

### Step 6: Substitute \( a \) in the eccentricity formula
The eccentricity \( e \) of the ellipse is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting \( a = 2b \):
\[ e = \sqrt{1 - \frac{b^2}{(2b)^2}} = \sqrt{1 - \frac{b^2}{4b^2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} \]

### Step 7: Final calculation of eccentricity
Thus, we find:
\[ e = \frac{\sqrt{3}}{2} \]

### Conclusion
The eccentricity of the ellipse is:
\[ \boxed{\frac{\sqrt{3}}{2}} \]