The length of the conjugate axis of the hyperbola ` 9x^(2) -16y^(2) - 18x -64y + 89 =0` is a) 8 b) 6 c) 4 d) 5

Options

A. 8
B. 6
C. 4
D. 5

Answer

To find the length of the conjugate axis of the hyperbola given by the equation \( 9x^2 - 16y^2 - 18x - 64y + 89 = 0 \), we will follow these steps:

### Step 1: Rearranging the Equation
Start with the given equation:
\[ 9x^2 - 16y^2 - 18x - 64y + 89 = 0 \]
Rearranging gives:
\[ 9x^2 - 18x - 16y^2 - 64y + 89 = 0 \]

### Step 2: Grouping Terms
Group the \(x\) and \(y\) terms:
\[ 9(x^2 - 2x) - 16(y^2 + 4y) + 89 = 0 \]

### Step 3: Completing the Square for \(x\)
To complete the square for \(x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]
Thus,
\[ 9((x - 1)^2 - 1) = 9(x - 1)^2 - 9 \]

### Step 4: Completing the Square for \(y\)
To complete the square for \(y\):
\[ y^2 + 4y = (y + 2)^2 - 4 \]
Thus,
\[ -16((y + 2)^2 - 4) = -16(y + 2)^2 + 64 \]

### Step 5: Substitute Back
Substituting back into the equation gives:
\[ 9((x - 1)^2 - 1) - 16((y + 2)^2 - 4) + 89 = 0 \]
This simplifies to:
\[ 9(x - 1)^2 - 9 - 16(y + 2)^2 + 64 + 89 = 0 \]
Combining constants:
\[ 9(x - 1)^2 - 16(y + 2)^2 + 144 = 0 \]
Rearranging gives:
\[ 9(x - 1)^2 - 16(y + 2)^2 = -144 \]

### Step 6: Dividing by -144
Dividing the entire equation by -144:
\[ -\frac{9(x - 1)^2}{144} + \frac{16(y + 2)^2}{144} = 1 \]
This simplifies to:
\[ \frac{(y + 2)^2}{9} - \frac{(x - 1)^2}{16} = 1 \]

### Step 7: Identifying Parameters
Now, we can compare this with the standard form of the hyperbola:
\[ \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 \]
From our equation, we have:
- \(a^2 = 9\) → \(a = 3\)
- \(b^2 = 16\) → \(b = 4\)

### Step 8: Finding the Length of the Conjugate Axis
The length of the conjugate axis is given by \(2b\):
\[ \text{Length of the conjugate axis} = 2b = 2 \times 3 = 6 \]

### Final Answer
Thus, the length of the conjugate axis of the hyperbola is:
\[ \boxed{6} \]