Find the eccentricity and coordinates of the foci of the curve `3x ^(2) - y ^(2) =4.`

Answer

To solve the problem of finding the eccentricity and coordinates of the foci of the hyperbola given by the equation \(3x^2 - y^2 = 4\), we will follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the given equation:
\[ 3x^2 - y^2 = 4 \]
To convert this into the standard form of a hyperbola, we divide every term by 4:
\[ \frac{3x^2}{4} - \frac{y^2}{4} = 1 \]
This can be rewritten as:
\[ \frac{x^2}{\frac{4}{3}} - \frac{y^2}{4} = 1 \]

### Step 2: Identify \(a^2\) and \(b^2\)
From the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), we can identify:
\[ a^2 = \frac{4}{3}, \quad b^2 = 4 \]

### Step 3: Calculate \(a\) and \(b\)
Now, we find \(a\) and \(b\):
\[ a = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}, \quad b = \sqrt{4} = 2 \]

### Step 4: Calculate \(c\) using the relationship \(c^2 = a^2 + b^2\)
For hyperbolas, the relationship between \(a\), \(b\), and \(c\) is given by:
\[ c^2 = a^2 + b^2 \]
Substituting the values we found:
\[ c^2 = \frac{4}{3} + 4 = \frac{4}{3} + \frac{12}{3} = \frac{16}{3} \]
Thus,
\[ c = \sqrt{\frac{16}{3}} = \frac{4}{\sqrt{3}} \]

### Step 5: Calculate the eccentricity \(e\)
The eccentricity \(e\) of a hyperbola is given by:
\[ e = \frac{c}{a} \]
Substituting the values of \(c\) and \(a\):
\[ e = \frac{\frac{4}{\sqrt{3}}}{\frac{2}{\sqrt{3}}} = \frac{4}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} = \frac{4}{2} = 2 \]

### Step 6: Find the coordinates of the foci
The coordinates of the foci for a hyperbola in the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) are given by:
\[ (0, \pm c) \]
Thus, the coordinates of the foci are:
\[ (0, \pm \frac{4}{\sqrt{3}}) \]

### Final Answer
- Eccentricity \(e = 2\)
- Coordinates of the foci: \((0, \frac{4}{\sqrt{3}})\) and \((0, -\frac{4}{\sqrt{3}})\)