Show that the equation `9x^2-16 y^2-18 x+32 y-151=0` represents a hyperbola.

Options

A. one of the directrix is `x=21//5`
B. the length of latus rectum is `9//2`
C. foci are (6, 1) and `(-4, 1)`
D. the eccentricity is 5/4

Explanation

To show that the equation \(9x^2 - 16y^2 - 18x + 32y - 151 = 0\) represents a hyperbola, we will manipulate the equation into a standard form.

### Step-by-Step Solution:

1. **Rearranging the Equation:**
Start with the original equation:
\[ 9x^2 - 16y^2 - 18x + 32y - 151 = 0 \]
Rearranging gives:
\[ 9x^2 - 18x - 16y^2 + 32y = 151 \]

2. **Completing the Square for \(x\):**
For the \(x\) terms \(9x^2 - 18x\):
- Factor out the 9:
\[ 9(x^2 - 2x) \]
- To complete the square, take half of the coefficient of \(x\) (which is -2), square it (getting 1), and add/subtract it inside the parentheses:
\[ 9(x^2 - 2x + 1 - 1) = 9((x - 1)^2 - 1) = 9(x - 1)^2 - 9 \]

3. **Completing the Square for \(y\):**
For the \(y\) terms \(-16y^2 + 32y\):
- Factor out -16:
\[ -16(y^2 - 2y) \]
- To complete the square, take half of the coefficient of \(y\) (which is -2), square it (getting 1), and add/subtract it:
\[ -16(y^2 - 2y + 1 - 1) = -16((y - 1)^2 - 1) = -16(y - 1)^2 + 16 \]

4. **Substituting Back:**
Substitute the completed squares back into the equation:
\[ 9((x - 1)^2 - 1) - 16((y - 1)^2 - 1) = 151 \]
Simplifying gives:
\[ 9(x - 1)^2 - 9 - 16(y - 1)^2 + 16 = 151 \]
Combine constants:
\[ 9(x - 1)^2 - 16(y - 1)^2 + 7 = 151 \]
Rearranging:
\[ 9(x - 1)^2 - 16(y - 1)^2 = 144 \]

5. **Dividing by 144:**
Divide the entire equation by 144 to normalize:
\[ \frac{9(x - 1)^2}{144} - \frac{16(y - 1)^2}{144} = 1 \]
Simplifying gives:
\[ \frac{(x - 1)^2}{16} - \frac{(y - 1)^2}{9} = 1 \]

6. **Identifying the Hyperbola:**
The equation now is in the standard form of a hyperbola:
\[ \frac{(x - 1)^2}{4^2} - \frac{(y - 1)^2}{3^2} = 1 \]
This confirms that the given equation represents a hyperbola.