What are the foci of the hyperbola `x^(2)/(36)-y^(2)/(16)=1`

Answer

To find the foci of the hyperbola given by the equation

\[ \frac{x^2}{36} - \frac{y^2}{16} = 1, \]

we can follow these steps:

### Step 1: Identify the values of \(a^2\) and \(b^2\)

From the equation of the hyperbola, we can identify:

\[ a^2 = 36 \quad \text{and} \quad b^2 = 16. \]

### Step 2: Calculate \(a\) and \(b\)

Taking the square roots, we find:

\[ a = \sqrt{36} = 6 \quad \text{and} \quad b = \sqrt{16} = 4. \]

### Step 3: Use the formula to find \(c\)

The relationship between \(a\), \(b\), and \(c\) (the distance from the center to each focus) for a hyperbola is given by:

\[ c^2 = a^2 + b^2. \]

Substituting the values we found:

\[ c^2 = 36 + 16 = 52. \]

### Step 4: Calculate \(c\)

Now, we take the square root to find \(c\):

\[ c = \sqrt{52} = \sqrt{4 \cdot 13} = 2\sqrt{13}. \]

### Step 5: Determine the coordinates of the foci

For a hyperbola of the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the foci are located at \((\pm c, 0)\). Therefore, the coordinates of the foci are:

\[ (\pm 2\sqrt{13}, 0). \]

### Final Answer

The foci of the hyperbola are:

\[ (2\sqrt{13}, 0) \quad \text{and} \quad (-2\sqrt{13}, 0). \]

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