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 hyperbola conjugate to the given hyperbola \(\frac{x^2}{4} - \frac{y^2}{12} = 1\).

### Step-by-step Solution:

1. **Identify the given hyperbola**:
The equation of the hyperbola is \(\frac{x^2}{4} - \frac{y^2}{12} = 1\). Here, we can identify \(a^2 = 4\) and \(b^2 = 12\).

2. **Calculate \(a\) and \(b\)**:
\[ a = \sqrt{4} = 2, \quad b = \sqrt{12} = 2\sqrt{3} \]

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

4. **Find the eccentricity \(e'\) of the conjugate hyperbola**:
For the conjugate hyperbola, the relationship between the eccentricities \(e\) and \(e'\) is given by:
\[ \frac{1}{e^2} + \frac{1}{e'^2} = 1 \]
We already found \(e = 2\), so:
\[ e^2 = 4 \]
Therefore:
\[ \frac{1}{4} + \frac{1}{e'^2} = 1 \]
Rearranging gives:
\[ \frac{1}{e'^2} = 1 - \frac{1}{4} = \frac{3}{4} \]
Thus:
\[ e'^2 = \frac{4}{3} \]

5. **Calculate \(3e'^2\)**:
Now we need to find \(3e'^2\):
\[ 3e'^2 = 3 \times \frac{4}{3} = 4 \]

### Final Answer:
\[ 3e^2 = 4 \]