If the eccentricity of the hyperbola conjugate to the hyperbola `(x^2)/(4)-(y^2)/(12)=1` is e, then `3e^2` is equal to:

Answer

To solve the problem, we need to find the value of \(3e^2\) where \(e\) is the eccentricity of the conjugate hyperbola of the given hyperbola \(\frac{x^2}{4} - \frac{y^2}{12} = 1\).

### Step-by-Step Solution:

1. **Identify the parameters of the given hyperbola:**
The equation of the hyperbola is given as:
\[ \frac{x^2}{4} - \frac{y^2}{12} = 1 \]
From this, we can identify \(a^2 = 4\) and \(b^2 = 12\). Thus, we have:
\[ a = 2 \quad \text{and} \quad b = \sqrt{12} = 2\sqrt{3} \]

2. **Calculate the eccentricity of the given hyperbola:**
The eccentricity \(e'\) of a hyperbola is given by the formula:
\[ e' = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting the values of \(a\) and \(b\):
\[ e' = \sqrt{1 + \frac{12}{4}} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

3. **Use the relationship between the eccentricities of the hyperbola and its conjugate:**
The relationship between the eccentricity \(e'\) of the hyperbola and the eccentricity \(e\) of its conjugate hyperbola is given by:
\[ \frac{1}{e'^2} + \frac{1}{e^2} = 1 \]
We already found \(e' = 2\), so we calculate \(e'^2\):
\[ e'^2 = 2^2 = 4 \]
Therefore:
\[ \frac{1}{4} + \frac{1}{e^2} = 1 \]

4. **Solve for \(e^2\):**
Rearranging the equation:
\[ \frac{1}{e^2} = 1 - \frac{1}{4} = \frac{3}{4} \]
Taking the reciprocal gives:
\[ e^2 = \frac{4}{3} \]

5. **Calculate \(3e^2\):**
We need to find \(3e^2\):
\[ 3e^2 = 3 \cdot \frac{4}{3} = 4 \]

### Final Answer:
Thus, the value of \(3e^2\) is \(4\).