If the distances of one focus of hyperbola from its directrices are `5` and `3`, then its eccentricity is

Options

A. `sqrt2`
B. 2
C. 4
D. 8

Explanation

To find the eccentricity of the hyperbola given the distances of one focus from its directrices, we can follow these steps:

### Step-by-step Solution:

1. **Understanding the Problem**: We know that the distances of one focus of the hyperbola from its directrices are given as 5 and 3. We denote these distances as \( d_1 = 5 \) and \( d_2 = 3 \).

2. **Using the Formula**: The distances of the focus from the directrices can be expressed in terms of the semi-major axis \( a \) and the eccentricity \( e \) of the hyperbola. The distances can be represented as:
\[ d_1 = ae + \frac{a}{e} \quad \text{(1)} \]
\[ d_2 = ae - \frac{a}{e} \quad \text{(2)} \]

3. **Setting Up the Equations**: From the given distances, we can set up the following equations:
\[ ae + \frac{a}{e} = 5 \quad \text{(1)} \]
\[ ae - \frac{a}{e} = 3 \quad \text{(2)} \]

4. **Adding the Equations**: We can add equations (1) and (2):
\[ (ae + \frac{a}{e}) + (ae - \frac{a}{e}) = 5 + 3 \]
This simplifies to:
\[ 2ae = 8 \]
Therefore, we find:
\[ ae = 4 \quad \text{(3)} \]

5. **Subtracting the Equations**: Now, we subtract equation (2) from equation (1):
\[ (ae + \frac{a}{e}) - (ae - \frac{a}{e}) = 5 - 3 \]
This simplifies to:
\[ 2\frac{a}{e} = 2 \]
Thus, we have:
\[ \frac{a}{e} = 1 \quad \text{(4)} \]

6. **Finding \( a \)**: From equation (3) \( ae = 4 \) and equation (4) \( \frac{a}{e} = 1 \), we can express \( a \) in terms of \( e \):
\[ a = e \]
Substituting \( a = e \) into equation (3):
\[ e \cdot e = 4 \]
This gives:
\[ e^2 = 4 \]
Therefore:
\[ e = 2 \]

7. **Conclusion**: The eccentricity \( e \) of the hyperbola is \( 2 \).

### Final Answer:
The eccentricity of the hyperbola is \( 2 \).