The vertices of the hyperbola `9x^(2)-16y^(2)-36x+96y-252=0` are

Options

A. `(6,3)` and `(-6,3)`
B. `(6,3)` and `(-2,3)`
C. `(-6,3)` and `(-6,-3)`
D. None of these

Explanation

To find the vertices of the hyperbola given by the equation \( 9x^2 - 16y^2 - 36x + 96y - 252 = 0 \), we will follow these steps:

### Step 1: Rearrange the equation
We start with the equation:
\[ 9x^2 - 16y^2 - 36x + 96y - 252 = 0 \]
Rearranging gives us:
\[ 9x^2 - 36x - 16y^2 + 96y = 252 \]

### Step 2: Complete the square for the \(x\) terms
For the \(x\) terms \(9x^2 - 36x\):
1. Factor out the 9:
\[ 9(x^2 - 4x) \]
2. To complete the square, take half of \(-4\) (which is \(-2\)), square it (getting \(4\)), and add/subtract it inside the parentheses:
\[ 9(x^2 - 4x + 4 - 4) = 9((x - 2)^2 - 4) = 9(x - 2)^2 - 36 \]

### Step 3: Complete the square for the \(y\) terms
For the \(y\) terms \(-16y^2 + 96y\):
1. Factor out \(-16\):
\[ -16(y^2 - 6y) \]
2. To complete the square, take half of \(-6\) (which is \(-3\)), square it (getting \(9\)), and add/subtract it inside the parentheses:
\[ -16(y^2 - 6y + 9 - 9) = -16((y - 3)^2 - 9) = -16(y - 3)^2 + 144 \]

### Step 4: Substitute back into the equation
Now substituting back into the rearranged equation:
\[ 9((x - 2)^2 - 4) - 16((y - 3)^2 - 9) = 252 \]
This simplifies to:
\[ 9(x - 2)^2 - 36 - 16(y - 3)^2 + 144 = 252 \]
Combining constants:
\[ 9(x - 2)^2 - 16(y - 3)^2 + 108 = 252 \]
Subtracting 108 from both sides:
\[ 9(x - 2)^2 - 16(y - 3)^2 = 144 \]

### Step 5: Divide by 144
Dividing the entire equation by 144 gives:
\[ \frac{(x - 2)^2}{16} - \frac{(y - 3)^2}{9} = 1 \]

### Step 6: Identify the center and vertices
The standard form of a hyperbola is:
\[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \]
From our equation, we have:
- Center: \((h, k) = (2, 3)\)
- \(a^2 = 16 \Rightarrow a = 4\)
- \(b^2 = 9 \Rightarrow b = 3\)

The vertices of the hyperbola are given by:
\[ (h \pm a, k) = (2 \pm 4, 3) \]
Calculating the vertices:
1. \( (2 + 4, 3) = (6, 3) \)
2. \( (2 - 4, 3) = (-2, 3) \)

### Final Answer:
The vertices of the hyperbola are:
\[ (6, 3) \text{ and } (-2, 3) \]