Find the axes, vertices, foci, eccentricity, equations of the directrices, and length of the latus rectum of the hyperbola `9x ^(2) - 16 y ^(2) = 144.`

Answer

To solve the problem, we will follow these steps:

### Step 1: Convert the equation to standard form
The given equation of the hyperbola is:
\[ 9x^2 - 16y^2 = 144 \]

First, we divide the entire equation by 144 to convert it into standard form:
\[ \frac{9x^2}{144} - \frac{16y^2}{144} = 1 \]
This simplifies to:
\[ \frac{x^2}{16} - \frac{y^2}{9} = 1 \]

### Step 2: Identify \(a^2\) and \(b^2\)
From the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), we can identify:
- \(a^2 = 16\) → \(a = \sqrt{16} = 4\)
- \(b^2 = 9\) → \(b = \sqrt{9} = 3\)

### Step 3: Find the lengths of the axes
The transverse axis length is given by \(2a\) and the conjugate axis length is given by \(2b\):
- Transverse axis = \(2a = 2 \times 4 = 8\)
- Conjugate axis = \(2b = 2 \times 3 = 6\)

### Step 4: Find the vertices
The vertices of the hyperbola are located at \((\pm a, 0)\):
- Vertices = \((\pm 4, 0)\) → \((4, 0)\) and \((-4, 0)\)

### Step 5: Find the eccentricity
The eccentricity \(e\) of the hyperbola is given by the formula:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4} \]

### Step 6: Find the foci
The foci are located at \((\pm ae, 0)\):
- Foci = \((\pm 4 \cdot \frac{5}{4}, 0) = (\pm 5, 0)\) → \((5, 0)\) and \((-5, 0)\)

### Step 7: Find the equations of the directrices
The equations of the directrices are given by:
\[ x = \pm \frac{a}{e} = \pm \frac{4}{\frac{5}{4}} = \pm \frac{16}{5} \]
Thus, the equations of the directrices are:
\[ x = \frac{16}{5} \quad \text{and} \quad x = -\frac{16}{5} \]

### Step 8: Find the length of the latus rectum
The length of the latus rectum \(L\) is given by:
\[ L = \frac{2b^2}{a} = \frac{2 \cdot 9}{4} = \frac{18}{4} = \frac{9}{2} \]

### Summary of Results
- **Transverse Axis Length**: 8
- **Conjugate Axis Length**: 6
- **Vertices**: (4, 0) and (-4, 0)
- **Foci**: (5, 0) and (-5, 0)
- **Eccentricity**: \(\frac{5}{4}\)
- **Equations of Directrices**: \(x = \frac{16}{5}\) and \(x = -\frac{16}{5}\)
- **Length of Latus Rectum**: \(\frac{9}{2}\)