If the distance between foci of a hyperbola is twice the distance between its directrices, then the eccentricity of conjugate hyperbola is :

Options

A. `2`
B. `sqrt(3)`
C. `sqrt(2)`
D. `3/2`

Explanation

To solve the problem, we need to find the eccentricity of the conjugate hyperbola given that the distance between the foci of a hyperbola is twice the distance between its directrices.

### Step-by-step Solution:

1. **Understand the Definitions**:
- The distance between the foci of a hyperbola is given by \(2Ae\), where \(A\) is the semi-major axis and \(e\) is the eccentricity of the hyperbola.
- The distance between the directrices of a hyperbola is given by \(\frac{2A}{e}\).

2. **Set Up the Equation**:
- According to the problem, the distance between the foci is twice the distance between the directrices:
\[ 2Ae = 2 \left(\frac{2A}{e}\right) \]

3. **Simplify the Equation**:
- We can simplify the equation:
\[ 2Ae = \frac{4A}{e} \]
- Cancel \(2A\) from both sides (assuming \(A \neq 0\)):
\[ e = \frac{4}{2} \cdot \frac{1}{e} \]
- This simplifies to:
\[ e^2 = 2 \]

4. **Find the Eccentricity of the Hyperbola**:
- Taking the square root of both sides gives:
\[ e = \sqrt{2} \]

5. **Use the Relationship Between Eccentricities**:
- For a hyperbola and its conjugate hyperbola, the relationship is given by:
\[ \frac{1}{e^2} + \frac{1}{e_c^2} = 1 \]
where \(e_c\) is the eccentricity of the conjugate hyperbola.

6. **Substitute the Known Value**:
- Substitute \(e^2 = 2\) into the equation:
\[ \frac{1}{2} + \frac{1}{e_c^2} = 1 \]

7. **Solve for the Eccentricity of the Conjugate Hyperbola**:
- Rearranging gives:
\[ \frac{1}{e_c^2} = 1 - \frac{1}{2} = \frac{1}{2} \]
- Taking the reciprocal:
\[ e_c^2 = 2 \]
- Therefore, taking the square root:
\[ e_c = \sqrt{2} \]

### Final Answer:
The eccentricity of the conjugate hyperbola is \(\sqrt{2}\).