Find the centre, eccentrcity, foci and directrices of the hyperboal :
`9x^(2) - 16y^(2) + 18x + 32y - 151 = 0 `.

Answer

To solve the problem of finding the center, eccentricity, foci, and directrices of the hyperbola given by the equation:

\[ 9x^{2} - 16y^{2} + 18x + 32y - 151 = 0 \]

we will follow these steps:

### Step 1: Rearranging the equation
First, we rearrange the equation to group the \(x\) and \(y\) terms together:

\[ 9x^{2} + 18x - 16y^{2} + 32y - 151 = 0 \]

### Step 2: Completing the square for \(x\) and \(y\)
Next, we will complete the square for the \(x\) terms and the \(y\) terms.

For \(x\):
\[ 9(x^{2} + 2x) = 9((x + 1)^{2} - 1) = 9(x + 1)^{2} - 9 \]

For \(y\):
\[ -16(y^{2} - 2y) = -16((y - 1)^{2} - 1) = -16(y - 1)^{2} + 16 \]

Substituting these back into the equation gives:

\[ 9(x + 1)^{2} - 9 - 16(y - 1)^{2} + 16 - 151 = 0 \]

Simplifying this, we have:

\[ 9(x + 1)^{2} - 16(y - 1)^{2} - 144 = 0 \]

### Step 3: Setting the equation to standard form
Now we can set the equation to standard form:

\[ 9(x + 1)^{2} - 16(y - 1)^{2} = 144 \]

Dividing through by 144 gives:

\[ \frac{(x + 1)^{2}}{16} - \frac{(y - 1)^{2}}{9} = 1 \]

### Step 4: Identifying parameters
From the standard form of the hyperbola:

\[ \frac{(x - h)^{2}}{a^{2}} - \frac{(y - k)^{2}}{b^{2}} = 1 \]

we can identify:
- Center \((h, k) = (-1, 1)\)
- \(a^{2} = 16 \Rightarrow a = 4\)
- \(b^{2} = 9 \Rightarrow b = 3\)

### Step 5: Finding the eccentricity
The eccentricity \(e\) of a hyperbola is given by:

\[ e = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4} \]

### Step 6: Finding the foci
The foci of the hyperbola are located at:

\[ (h \pm c, k) \text{ where } c = \sqrt{a^{2} + b^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

Thus, the foci are:

\[ (-1 \pm 5, 1) \Rightarrow (4, 1) \text{ and } (-6, 1) \]

### Step 7: Finding the directrices
The equations of the directrices are given by:

\[ x = h \pm \frac{a}{e} \]

Calculating \(\frac{a}{e} = \frac{4}{\frac{5}{4}} = \frac{16}{5}\), we have:

\[ x = -1 \pm \frac{16}{5} \]

Calculating these gives:

\[ x = -1 + \frac{16}{5} = \frac{11}{5} \quad \text{and} \quad x = -1 - \frac{16}{5} = -\frac{21}{5} \]

### Final Answers
- **Center**: \((-1, 1)\)
- **Eccentricity**: \(\frac{5}{4}\)
- **Foci**: \((4, 1)\) and \((-6, 1)\)
- **Directrices**: \(x = \frac{11}{5}\) and \(x = -\frac{21}{5}\)