The vertices of the hyperbola ` ((x-2)^(2))/( 9) -((y-3)^(2))/( 4) =1` are

Options

A. (2,3), (-1,3)
B. (5,3), (-1,3)
C. (0,3), (4,3)
D. (1,3), (3,3)

Answer

To find the vertices of the hyperbola given by the equation

\[ \frac{(x-2)^2}{9} - \frac{(y-3)^2}{4} = 1, \]

we can follow these steps:

### Step 1: Identify the standard form of the hyperbola
The given equation is already in the standard form of a hyperbola, which is

\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1, \]

where \((h, k)\) is the center of the hyperbola, \(a^2\) is the denominator of the \(x\) term, and \(b^2\) is the denominator of the \(y\) term.

### Step 2: Determine the center of the hyperbola
From the equation, we can see that:
- \(h = 2\)
- \(k = 3\)

Thus, the center of the hyperbola is at the point \((2, 3)\).

### Step 3: Find the values of \(a\) and \(b\)
From the equation:
- \(a^2 = 9\) implies \(a = \sqrt{9} = 3\)
- \(b^2 = 4\) implies \(b = \sqrt{4} = 2\)

### Step 4: Calculate the vertices
For a hyperbola of the form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), the vertices are located at:
- \((h-a, k)\) and \((h+a, k)\)

Substituting the values we found:
- The first vertex is \((2-3, 3) = (-1, 3)\)
- The second vertex is \((2+3, 3) = (5, 3)\)

### Step 5: State the vertices
Thus, the vertices of the hyperbola are:
\[ (-1, 3) \text{ and } (5, 3). \]

### Final Answer
The vertices of the hyperbola are \((-1, 3)\) and \((5, 3)\).

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