The foci of the hyperbola `4x^(2)-9y^(2)-36=0` are

Options

A. `[pmsqrt((11)),0]`
B. `[pmsqrt((12)),0]`
C. `[pmsqrt((13)),0]`
D. `[0,pmsqrt((12))]`

Answer

To find the foci of the hyperbola given by the equation \(4x^2 - 9y^2 - 36 = 0\), we will follow these steps:

### Step 1: Rewrite the equation in standard form
Start with the given equation:
\[ 4x^2 - 9y^2 - 36 = 0 \]
Rearranging gives:
\[ 4x^2 - 9y^2 = 36 \]
Now, divide the entire equation by 36:
\[ \frac{4x^2}{36} - \frac{9y^2}{36} = 1 \]
This simplifies to:
\[ \frac{x^2}{9} - \frac{y^2}{4} = 1 \]

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

### Step 3: 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 \(a\) and \(b\):
\[ e = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{9 + 4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3} \]

### Step 4: Find the coordinates of the foci
The foci of the hyperbola are located at \((\pm ae, 0)\). Therefore, we calculate:
\[ ae = 3 \cdot \frac{\sqrt{13}}{3} = \sqrt{13} \]
Thus, the coordinates of the foci are:
\[ (\sqrt{13}, 0) \quad \text{and} \quad (-\sqrt{13}, 0) \]

### Final Result
The foci of the hyperbola \(4x^2 - 9y^2 - 36 = 0\) are:
\[ (\sqrt{13}, 0) \quad \text{and} \quad (-\sqrt{13}, 0) \]

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