The foci of the hyperbola `4x^(2)-9y^(2)-1=0` are:

Options

A. `(pm sqrt(3),0)`
B. `(pm (sqrt(13))/(6),0)`
C. `(0, pm (sqrt(13))/(6))`
D. None of these

Answer

To find the foci of the hyperbola given by the equation \(4x^2 - 9y^2 - 1 = 0\), we will follow these steps:

### Step 1: Rewrite the equation in standard form
Start by rearranging the equation to isolate the constant on one side:
\[ 4x^2 - 9y^2 = 1 \]

### Step 2: Divide by 1
Next, divide the entire equation by 1 to express it in the standard form of a hyperbola:
\[ \frac{x^2}{\frac{1}{4}} - \frac{y^2}{\frac{1}{9}} = 1 \]

### Step 3: Identify \(a^2\) and \(b^2\)
From the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), we can identify:
\[ a^2 = \frac{1}{4} \quad \text{and} \quad b^2 = \frac{1}{9} \]
Thus, we have:
\[ a = \frac{1}{2} \quad \text{and} \quad b = \frac{1}{3} \]

### Step 4: Calculate the eccentricity \(e\)
The eccentricity \(e\) of a hyperbola is given by the formula:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting the values of \(a^2\) and \(b^2\):
\[ e = \sqrt{1 + \frac{\frac{1}{9}}{\frac{1}{4}}} = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3} \]

### Step 5: Find the coordinates of the foci
The coordinates of the foci for a hyperbola centered at the origin with a horizontal transverse axis are given by:
\[ (\pm ae, 0) \]
Substituting the values of \(a\) and \(e\):
\[ ae = \frac{1}{2} \cdot \frac{\sqrt{13}}{3} = \frac{\sqrt{13}}{6} \]
Thus, the coordinates of the foci are:
\[ \left(\pm \frac{\sqrt{13}}{6}, 0\right) \]

### Final Answer
The foci of the hyperbola \(4x^2 - 9y^2 - 1 = 0\) are:
\[ \left(\frac{\sqrt{13}}{6}, 0\right) \text{ and } \left(-\frac{\sqrt{13}}{6}, 0\right) \]