What is the eccentricity of an ellipse, if its latus rectum is equal to one - half of its minor axis?

Options

A. `1//4`
B. `1//2`
C. `sqrt3//4`
D. `sqrt3//2`

Answer

To find the eccentricity of an ellipse given that its latus rectum is equal to one-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 **minor axis (M)** of the ellipse is given by:
\[ M = 2b \]

### Step 2: Set up the equation
According to the problem, the latus rectum is equal to one-half of the minor axis:
\[ L = \frac{1}{2} M \]
Substituting the expressions for \(L\) and \(M\):
\[ \frac{2b^2}{a} = \frac{1}{2} \cdot 2b \]
This simplifies to:
\[ \frac{2b^2}{a} = b \]

### Step 3: Solve for \(b\) in terms of \(a\)
To eliminate \(b\) from the equation, we can multiply both sides by \(a\):
\[ 2b^2 = ab \]
Assuming \(b \neq 0\), we can divide both sides by \(b\):
\[ 2b = a \]
This gives us:
\[ a = 2b \]

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

---