The length of the latus rectum of an ellipse is `1/3` of the major axis. Its eccentricity is

Options

A. `2/3`
B. `sqrt((2)/(3))`
C. `(5xx4xx3)/(7 ^(3))`
D. `((3)/(4)) ^(4)`

Answer

To solve the problem, we need to find the eccentricity of the ellipse given that the length of the latus rectum is \( \frac{1}{3} \) of the major axis.

### Step 1: Define the terms
Let the length of the major axis be \( 2a \), where \( a \) is the semi-major axis. Therefore, the length of the major axis is \( 2a \).

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

### Step 3: Relate the latus rectum to the major axis
According to the problem, the length of the latus rectum is \( \frac{1}{3} \) of the major axis:
\[ \frac{2b^2}{a} = \frac{1}{3} \cdot 2a \]
This simplifies to:
\[ \frac{2b^2}{a} = \frac{2a}{3} \]

### Step 4: Simplify the equation
We can cancel \( 2 \) from both sides:
\[ \frac{b^2}{a} = \frac{a}{3} \]
Multiplying both sides by \( 3a \) gives:
\[ 3b^2 = a^2 \]

### Step 5: Use the relationship between \( a \), \( b \), and eccentricity
The relationship between \( a \), \( b \), and the eccentricity \( e \) of an ellipse is given by:
\[ b^2 = a^2(1 - e^2) \]
Substituting \( b^2 \) from the previous step:
\[ 3b^2 = 3a^2(1 - e^2) = a^2 \]
This simplifies to:
\[ 3(1 - e^2) = 1 \]

### Step 6: Solve for \( e^2 \)
Rearranging gives:
\[ 3 - 3e^2 = 1 \]
\[ 3e^2 = 2 \]
\[ e^2 = \frac{2}{3} \]

### Step 7: Find \( e \)
Taking the square root gives us the eccentricity:
\[ e = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{6}}{3} \]

### Final Answer
The eccentricity \( e \) of the ellipse is:
\[ e = \frac{\sqrt{6}}{3} \]