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: Understand the definitions
The latus rectum (L) of an ellipse is given by the formula:
\[ L = \frac{2b^2}{a} \]
where \(a\) is the semi-major axis and \(b\) is the semi-minor axis. The length of the minor axis is \(2b\).

### Step 2: Set up the equation
According to the problem, the latus rectum is half of the minor axis. Therefore, we can write:
\[ \frac{2b^2}{a} = \frac{1}{2} \times 2b \]
This simplifies to:
\[ \frac{2b^2}{a} = b \]

### Step 3: Rearrange the equation
To eliminate \(b\) from the equation, we can multiply both sides by \(a\):
\[ 2b^2 = ab \]
Now, rearranging gives:
\[ 2b^2 - ab = 0 \]

### Step 4: Factor the equation
We can factor out \(b\):
\[ b(2b - a) = 0 \]
Since \(b\) cannot be zero (as it represents the semi-minor axis), we have:
\[ 2b - a = 0 \quad \Rightarrow \quad a = 2b \]

### Step 5: Use the eccentricity formula
The eccentricity \(e\) of an ellipse is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting \(a = 2b\) into the formula:
\[ e = \sqrt{1 - \frac{b^2}{(2b)^2}} = \sqrt{1 - \frac{b^2}{4b^2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

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

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