The foci of the hyperbola `(x^(2))/(16) -(y^(2))/(9) = 1 ` is :

Options

A. (0,5)
B. (5,0)
C. `(pm 5, 0) `
D. `(0, pm 5)`

Answer

To find the foci of the hyperbola given by the equation \(\frac{x^2}{16} - \frac{y^2}{9} = 1\), we can follow these steps:

### Step 1: Identify the values of \(a\) and \(b\)
The standard form of a hyperbola is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
Comparing this with the given equation \(\frac{x^2}{16} - \frac{y^2}{9} = 1\), we can identify:
\[ a^2 = 16 \quad \text{and} \quad b^2 = 9 \]
Thus, we find:
\[ a = \sqrt{16} = 4 \quad \text{and} \quad b = \sqrt{9} = 3 \]

### Step 2: Calculate the value of \(c\)
For hyperbolas, the relationship between \(a\), \(b\), and \(c\) (the distance from the center to the foci) is given by:
\[ c^2 = a^2 + b^2 \]
Substituting the values of \(a^2\) and \(b^2\):
\[ c^2 = 16 + 9 = 25 \]
Taking the square root gives:
\[ c = \sqrt{25} = 5 \]

### Step 3: Determine the coordinates of the foci
The foci of a hyperbola in the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) are located at \((\pm c, 0)\). Therefore, the coordinates of the foci are:
\[ (\pm 5, 0) \]

### Final Answer
The foci of the hyperbola \(\frac{x^2}{16} - \frac{y^2}{9} = 1\) are:
\[ (5, 0) \quad \text{and} \quad (-5, 0) \]

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