If the distance between the foci of an ellipse is 8 and length of latus rectum is `18/5,` then the eccentricity of ellipse is:

Options

A. `2/5`
B. `3/5`
C. `4/5`
D. None of these

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We are given:
- The distance between the foci of the ellipse is 8.
- The length of the latus rectum is \( \frac{18}{5} \).

### Step 2: Relate the distance between the foci to the parameters of the ellipse
The distance between the foci of an ellipse is given by the formula:
\[ 2ae = 8 \]
From this, we can find \( ae \):
\[ ae = \frac{8}{2} = 4 \]

### Step 3: Use the formula for the length of the latus rectum
The length of the latus rectum \( L \) is given by:
\[ L = \frac{2b^2}{a} \]
We know that \( L = \frac{18}{5} \), so we can set up the equation:
\[ \frac{2b^2}{a} = \frac{18}{5} \]
From this, we can express \( b^2 \) in terms of \( a \):
\[ 2b^2 = \frac{18}{5}a \implies b^2 = \frac{9}{5}a \]

### Step 4: Substitute \( b^2 \) into the equation for \( a^2 + b^2 \)
We know from the earlier step that:
\[ a^2 + b^2 = 16 \]
Substituting \( b^2 = \frac{9}{5}a \) into this equation gives:
\[ a^2 + \frac{9}{5}a = 16 \]

### Step 5: Clear the fraction by multiplying through by 5
To eliminate the fraction, multiply the entire equation by 5:
\[ 5a^2 + 9a = 80 \]

### Step 6: Rearrange into standard quadratic form
Rearranging gives:
\[ 5a^2 + 9a - 80 = 0 \]

### Step 7: Factor or use the quadratic formula to solve for \( a \)
We can factor this quadratic equation:
\[ (5a - 16)(a + 5) = 0 \]
Setting each factor to zero gives:
\[ 5a - 16 = 0 \quad \text{or} \quad a + 5 = 0 \]
Thus,
\[ a = \frac{16}{5} \quad \text{or} \quad a = -5 \]
Since \( a \) must be positive, we take \( a = 5 \).

### Step 8: Find \( b^2 \)
Using \( a = 5 \) in the equation \( b^2 = \frac{9}{5}a \):
\[ b^2 = \frac{9}{5} \cdot 5 = 9 \]

### Step 9: Calculate the eccentricity \( e \)
The eccentricity \( e \) is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

### Final Answer
Thus, the eccentricity of the ellipse is:
\[ \boxed{\frac{4}{5}} \]