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

Options

A. (2, 3), (5, 7)
B. (4, 1) (-6, 1)
C. (0, 0), (5, 3)
D. None

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: Rearrange the equation
Start by rearranging the equation to group the \(x\) and \(y\) terms:
\[ 9x^2 + 18x - 16y^2 + 32y - 151 = 0 \]

### Step 2: Complete the square for \(x\)
To complete the square for the \(x\) terms \(9x^2 + 18x\):
1. Factor out the coefficient of \(x^2\):
\[ 9(x^2 + 2x) \]
2. Complete the square:
\[ x^2 + 2x = (x + 1)^2 - 1 \]
3. Substitute back:
\[ 9((x + 1)^2 - 1) = 9(x + 1)^2 - 9 \]

### Step 3: Complete the square for \(y\)
Now complete the square for the \(y\) terms \(-16y^2 + 32y\):
1. Factor out \(-16\):
\[ -16(y^2 - 2y) \]
2. Complete the square:
\[ y^2 - 2y = (y - 1)^2 - 1 \]
3. Substitute back:
\[ -16((y - 1)^2 - 1) = -16(y - 1)^2 + 16 \]

### Step 4: Substitute back into the equation
Now substitute the completed squares back into the equation:
\[ 9((x + 1)^2 - 1) - 16((y - 1)^2 - 1) - 151 = 0 \]
This simplifies to:
\[ 9(x + 1)^2 - 9 - 16(y - 1)^2 + 16 - 151 = 0 \]
Combine the constants:
\[ 9(x + 1)^2 - 16(y - 1)^2 - 144 = 0 \]

### Step 5: Rearrange to standard form
Rearranging gives:
\[ 9(x + 1)^2 - 16(y - 1)^2 = 144 \]
Dividing by 144:
\[ \frac{(x + 1)^2}{16} - \frac{(y - 1)^2}{9} = 1 \]

### Step 6: Identify parameters
From the standard form \(\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\), we have:
- Center \((h, k) = (-1, 1)\)
- \(a^2 = 16 \Rightarrow a = 4\)
- \(b^2 = 9 \Rightarrow b = 3\)

### Step 7: Calculate \(c\)
Using the relationship \(c^2 = a^2 + b^2\):
\[ c^2 = 16 + 9 = 25 \Rightarrow c = 5 \]

### Step 8: Find the foci
The foci are located at \((h \pm c, k)\):
1. Foci coordinates:
\[ (-1 + 5, 1) = (4, 1) \]
\[ (-1 - 5, 1) = (-6, 1) \]

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

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