Find the equation of the hyperbola whose vertices are `(pm3,0)` and foci at `(pm5,0)`

Answer

To find the equation of the hyperbola with given vertices and foci, we can follow these steps:

### Step 1: Identify the center, vertices, and foci
The vertices are given at \((\pm 3, 0)\) and the foci at \((\pm 5, 0)\). The center of the hyperbola is at the origin \((0, 0)\).

### Step 2: Determine the values of \(a\) and \(c\)
- The distance from the center to each vertex is denoted as \(a\). Here, \(a = 3\).
- The distance from the center to each focus is denoted as \(c\). Here, \(c = 5\).

### Step 3: Use the relationship between \(a\), \(b\), and \(c\)
For hyperbolas, the relationship between \(a\), \(b\), and \(c\) is given by the equation:
\[ c^2 = a^2 + b^2 \]
We can rearrange this to find \(b^2\):
\[ b^2 = c^2 - a^2 \]

### Step 4: Calculate \(b^2\)
Substituting the values of \(a\) and \(c\):
\[ c^2 = 5^2 = 25 \]
\[ a^2 = 3^2 = 9 \]
Now, substituting these into the equation for \(b^2\):
\[ b^2 = 25 - 9 = 16 \]

### Step 5: Write the standard form of the hyperbola
The standard form of the equation of a hyperbola that opens horizontally is:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
Substituting \(a^2\) and \(b^2\) into this equation:
\[ \frac{x^2}{9} - \frac{y^2}{16} = 1 \]

### Step 6: Clear the denominators
To express the equation in a more standard form, we can multiply through by the least common multiple of the denominators (which is 144):
\[ 144 \left(\frac{x^2}{9} - \frac{y^2}{16}\right) = 144 \]
This simplifies to:
\[ 16x^2 - 9y^2 = 144 \]

### Final Result
Thus, the equation of the hyperbola is:
\[ \boxed{16x^2 - 9y^2 = 144} \]