Find the (i)length of the axis(ii)coordinate of vertices (iii)coordinate of the focii (iv)ecentricity(v)length of latus rectum
`(x^(2))/(9)-(y^(2))/(16)=1`

Answer

To solve the problem step by step, we will analyze the hyperbola given by the equation \(\frac{x^2}{9} - \frac{y^2}{16} = 1\).

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

### Step 2: Find the length of the transverse axis
The length of the transverse axis of a hyperbola is given by:
\[ \text{Length of transverse axis} = 2a \]
Substituting the value of \(a\):
\[ \text{Length of transverse axis} = 2 \times 3 = 6 \]

### Step 3: Find the coordinates of the vertices
The vertices of the hyperbola are located at \((\pm a, 0)\). Thus:
\[ \text{Coordinates of the vertices} = (3, 0) \text{ and } (-3, 0) \]

### Step 4: Find the coordinates of the foci
The distance \(c\) from the center to each focus is given by:
\[ c = \sqrt{a^2 + b^2} \]
Calculating \(c\):
\[ c = \sqrt{9 + 16} = \sqrt{25} = 5 \]
The coordinates of the foci are located at \((\pm c, 0)\):
\[ \text{Coordinates of the foci} = (5, 0) \text{ and } (-5, 0) \]

### Step 5: Find the eccentricity
The eccentricity \(e\) of the hyperbola is given by:
\[ e = \frac{c}{a} \]
Substituting the values of \(c\) and \(a\):
\[ e = \frac{5}{3} \]

### Step 6: Find the length of the latus rectum
The length of the latus rectum \(L\) is given by:
\[ L = \frac{2b^2}{a} \]
Calculating \(L\):
\[ L = \frac{2 \times 16}{3} = \frac{32}{3} \]

### Summary of Results
1. Length of the transverse axis: \(6\)
2. Coordinates of the vertices: \((3, 0)\) and \((-3, 0)\)
3. Coordinates of the foci: \((5, 0)\) and \((-5, 0)\)
4. Eccentricity: \(\frac{5}{3}\)
5. Length of the latus rectum: \(\frac{32}{3}\)