Find the coordinate of the foci, coordinate of the vertices, eccentricity and the length of the latus rectum of the hyperbola
`16x ^(2) - 9y ^(2) =576`

Answer

To solve the problem, we need to analyze the hyperbola given by the equation \(16x^2 - 9y^2 = 576\). We will find the coordinates of the foci, the coordinates of the vertices, the eccentricity, and the length of the latus rectum step by step.

### Step 1: Rewrite the equation in standard form
We start with the equation:
\[ 16x^2 - 9y^2 = 576 \]
To convert it to standard form, we divide the entire equation by 576:
\[ \frac{16x^2}{576} - \frac{9y^2}{576} = 1 \]
This simplifies to:
\[ \frac{x^2}{36} - \frac{y^2}{64} = 1 \]
Thus, we can identify \(a^2 = 36\) and \(b^2 = 64\).

### Step 2: Find the values of \(a\) and \(b\)
From the values of \(a^2\) and \(b^2\):
\[ a = \sqrt{36} = 6 \]
\[ b = \sqrt{64} = 8 \]

### Step 3: Calculate the eccentricity \(e\)
The eccentricity \(e\) of a hyperbola is given by the formula:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting the values of \(b^2\) and \(a^2\):
\[ e = \sqrt{1 + \frac{64}{36}} = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3} \]

### Step 4: Find the coordinates of the foci
The coordinates of the foci for a hyperbola are given by \((\pm ae, 0)\):
\[ ae = 6 \cdot \frac{5}{3} = 10 \]
Thus, the coordinates of the foci are:
\[ (10, 0) \quad \text{and} \quad (-10, 0) \]

### Step 5: Find the coordinates of the vertices
The coordinates of the vertices for a hyperbola are given by \((\pm a, 0)\):
\[ (6, 0) \quad \text{and} \quad (-6, 0) \]

### Step 6: Calculate the length of the latus rectum
The length of the latus rectum \(L\) is given by the formula:
\[ L = \frac{2b^2}{a} \]
Substituting the values of \(b^2\) and \(a\):
\[ L = \frac{2 \cdot 64}{6} = \frac{128}{6} = \frac{64}{3} \]

### Summary of Results
- **Coordinates of the foci:** \((10, 0)\) and \((-10, 0)\)
- **Coordinates of the vertices:** \((6, 0)\) and \((-6, 0)\)
- **Eccentricity:** \(\frac{5}{3}\)
- **Length of the latus rectum:** \(\frac{64}{3}\)