The length of the transverse axis of the hyperbola `9x^(2)-16y^(2)-18x -32y - 151 = 0` is

Options

A. 8
B. 2
C. 6
D. 2

Explanation

To find the length of the transverse axis 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
We start with the equation:
\[ 9x^2 - 16y^2 - 18x - 32y - 151 = 0 \]
We can rearrange this to group the \(x\) and \(y\) terms:
\[ 9x^2 - 18x - 16y^2 - 32y = 151 \]

### Step 2: Complete the square for the \(x\) terms
For the \(x\) terms \(9x^2 - 18x\), we factor out 9:
\[ 9(x^2 - 2x) \]
Now, we complete the square:
\[ x^2 - 2x = (x - 1)^2 - 1 \]
So,
\[ 9((x - 1)^2 - 1) = 9(x - 1)^2 - 9 \]

### Step 3: Complete the square for the \(y\) terms
For the \(y\) terms \(-16y^2 - 32y\), we factor out -16:
\[ -16(y^2 + 2y) \]
Now, we complete the square:
\[ y^2 + 2y = (y + 1)^2 - 1 \]
So,
\[ -16((y + 1)^2 - 1) = -16(y + 1)^2 + 16 \]

### Step 4: Substitute back into the equation
Now we substitute back into our rearranged equation:
\[ 9(x - 1)^2 - 9 - 16(y + 1)^2 + 16 = 151 \]
This simplifies to:
\[ 9(x - 1)^2 - 16(y + 1)^2 + 7 = 151 \]
Subtract 7 from both sides:
\[ 9(x - 1)^2 - 16(y + 1)^2 = 144 \]

### Step 5: Divide by 144 to get the standard form
Now, we divide the entire equation by 144:
\[ \frac{9(x - 1)^2}{144} - \frac{16(y + 1)^2}{144} = 1 \]
This simplifies to:
\[ \frac{(x - 1)^2}{16} - \frac{(y + 1)^2}{9} = 1 \]
This is now in the standard form of a hyperbola:
\[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \]
where \(h = 1\), \(k = -1\), \(a^2 = 16\) (thus \(a = 4\)), and \(b^2 = 9\) (thus \(b = 3\)).

### Step 6: Find the length of the transverse axis
The length of the transverse axis for a hyperbola is given by \(2a\):
\[ \text{Length of transverse axis} = 2a = 2 \times 4 = 8 \]

### Final Answer
The length of the transverse axis of the hyperbola is \(8\).

---