Find the eccentricity, length of a latus rectum, equations of the latus rectum of the hyperbola `(x^(2))/(16)-(y^(2))/(9) =1`.

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the values of \( a \) and \( b \)
The given hyperbola is:
\[ \frac{x^2}{16} - \frac{y^2}{9} = 1 \]
From the standard form of the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), we can identify:
- \( a^2 = 16 \) which gives \( a = \sqrt{16} = 4 \)
- \( b^2 = 9 \) which gives \( b = \sqrt{9} = 3 \)

### Step 2: Calculate the eccentricity \( e \)
The formula for the eccentricity \( e \) of a hyperbola is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting the values of \( a \) and \( b \):
\[ e = \sqrt{1 + \frac{9}{16}} = \sqrt{1 + 0.5625} = \sqrt{1.5625} = \frac{5}{4} \]

### Step 3: Calculate the length of the latus rectum
The length of the latus rectum \( L \) of a hyperbola is given by:
\[ L = \frac{2b^2}{a} \]
Substituting the values of \( b \) and \( a \):
\[ L = \frac{2 \times 9}{4} = \frac{18}{4} = \frac{9}{2} \]

### Step 4: Find the equations of the latus rectum
The equations of the latus rectum for a hyperbola centered at the origin are given by:
\[ y = \pm \frac{b^2}{a} \text{ at } x = \pm a \]
Substituting the values of \( a \) and \( b \):
\[ y = \pm \frac{9}{4} \text{ at } x = \pm 4 \]
Thus, the equations of the latus rectum are:
\[ x = 4, \quad y = \frac{9}{4} \quad \text{and} \quad y = -\frac{9}{4} \]
\[ x = -4, \quad y = \frac{9}{4} \quad \text{and} \quad y = -\frac{9}{4} \]

### Final Answers
- Eccentricity \( e = \frac{5}{4} \)
- Length of the latus rectum \( L = \frac{9}{2} \)
- Equations of the latus rectum:
- \( x = 4, y = \frac{9}{4} \)
- \( x = 4, y = -\frac{9}{4} \)
- \( x = -4, y = \frac{9}{4} \)
- \( x = -4, y = -\frac{9}{4} \)