The foci of the hyperbola ` 9x^(2) -16y ^(2)+18x +32 y- 151=0 ` are

Options

A. (2,1), (-6,1)
B. (-2,5), (-2,3)
C. (4,1), (-6,1)
D. (-2,4), (-2,-2)

Answer

To find the foci of the hyperbola given by the equation \( 9x^{2} - 16y^{2} + 18x + 32y - 151 = 0 \), we will follow these steps:

### Step 1: Rearranging the Equation
Start by rearranging the terms in the equation:
\[ 9x^{2} + 18x - 16y^{2} + 32y - 151 = 0 \]

### Step 2: Completing the Square
Next, we will complete the square for the \(x\) and \(y\) terms.

1. For \(x\):
\[ 9(x^{2} + 2x) = 9((x + 1)^{2} - 1) = 9(x + 1)^{2} - 9 \]

2. For \(y\):
\[ -16(y^{2} - 2y) = -16((y - 1)^{2} - 1) = -16(y - 1)^{2} + 16 \]

Substituting these back into the equation gives:
\[ 9((x + 1)^{2} - 1) - 16((y - 1)^{2} - 1) - 151 = 0 \]
This simplifies to:
\[ 9(x + 1)^{2} - 16(y - 1)^{2} - 9 + 16 - 151 = 0 \]
\[ 9(x + 1)^{2} - 16(y - 1)^{2} - 144 = 0 \]

### Step 3: Setting the Equation to Standard Form
Rearranging gives:
\[ 9(x + 1)^{2} - 16(y - 1)^{2} = 144 \]
Dividing the entire equation by 144 results in:
\[ \frac{(x + 1)^{2}}{16} - \frac{(y - 1)^{2}}{9} = 1 \]

### Step 4: Identifying Parameters
From the standard form of the hyperbola:
\[ \frac{(x - h)^{2}}{a^{2}} - \frac{(y - k)^{2}}{b^{2}} = 1 \]
we identify:
- \(h = -1\), \(k = 1\)
- \(a^{2} = 16 \Rightarrow a = 4\)
- \(b^{2} = 9 \Rightarrow b = 3\)

### Step 5: Finding the Foci
The foci of a hyperbola are given by the formula:
\[ c = \sqrt{a^{2} + b^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 \]
The coordinates of the foci are:
\[ (h \pm c, k) = (-1 \pm 5, 1) \]
This gives us the foci:
\[ (-1 + 5, 1) = (4, 1) \quad \text{and} \quad (-1 - 5, 1) = (-6, 1) \]

### Final Answer
The foci of the hyperbola are:
\[ (4, 1) \quad \text{and} \quad (-6, 1) \]

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