The vertices of a hyperbola are at (0, 0) and (10, 0) and one of its foci is at (18, 0). The equation of the hyperbola is

Options

A. `x^(2)/(25)-y^(2)/(144)=2`
B. `(x-5)^(2)/(25)-y^(2)/(144)=1`
C. `x^(2)/(25)-(y-5)^(2)/(144)=1`
D. `(x-5)^(2)/(25)-(y-5)^(2)/(144)=1`

Answer

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

### Step 1: Identify the vertices and the center
The vertices of the hyperbola are given as (0, 0) and (10, 0). The center of the hyperbola is the midpoint of the vertices.

**Calculation:**
\[ \text{Center} = \left( \frac{0 + 10}{2}, \frac{0 + 0}{2} \right) = (5, 0) \]

### Step 2: Find the value of \(a\)
The distance between the two vertices is equal to \(2a\).

**Calculation:**
\[ \text{Distance} = 10 - 0 = 10 \quad \Rightarrow \quad 2a = 10 \quad \Rightarrow \quad a = 5 \]

### Step 3: Identify the focus and find \(c\)
One of the foci is given at (18, 0). The distance from the center to the focus is denoted as \(c\).

**Calculation:**
\[ c = \text{Distance from center to focus} = 18 - 5 = 13 \]

### Step 4: Use the relationship between \(a\), \(b\), and \(c\)
For hyperbolas, the relationship between \(a\), \(b\), and \(c\) is given by:
\[ c^2 = a^2 + b^2 \]

**Calculation:**
\[ c^2 = 13^2 = 169 \]
\[ a^2 = 5^2 = 25 \]
Now substituting these values:
\[ 169 = 25 + b^2 \quad \Rightarrow \quad b^2 = 169 - 25 = 144 \]

### Step 5: Write the equation of the hyperbola
The standard form of the equation of a hyperbola centered at \((h, k)\) with a horizontal transverse axis is:
\[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \]
Here, \(h = 5\), \(k = 0\), \(a^2 = 25\), and \(b^2 = 144\).

**Final Equation:**
\[ \frac{(x - 5)^2}{25} - \frac{y^2}{144} = 1 \]

### Summary
The equation of the hyperbola is:
\[ \frac{(x - 5)^2}{25} - \frac{y^2}{144} = 1 \]

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