The vertices of the ellipse
`9x^(2)+4y^(2)-18x-27 =0` are

Options

A. `(1,+-2)`
B. `(1,+-3)`
C. `(1,+- 4)`
D. none of these

Explanation

To find the vertices of the ellipse given by the equation \(9x^2 + 4y^2 - 18x - 27 = 0\), we will follow these steps:

### Step 1: Rearrange the equation
We start with the equation:
\[ 9x^2 + 4y^2 - 18x - 27 = 0 \]
We can rearrange this to isolate the terms involving \(x\) and \(y\):
\[ 9x^2 - 18x + 4y^2 = 27 \]

### Step 2: Complete the square for the \(x\) terms
Next, we will complete the square for the \(x\) terms. We factor out the 9 from the \(x\) terms:
\[ 9(x^2 - 2x) + 4y^2 = 27 \]
Now, we complete the square for \(x^2 - 2x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]
Substituting this back, we have:
\[ 9((x - 1)^2 - 1) + 4y^2 = 27 \]
This simplifies to:
\[ 9(x - 1)^2 - 9 + 4y^2 = 27 \]
Adding 9 to both sides gives:
\[ 9(x - 1)^2 + 4y^2 = 36 \]

### Step 3: Divide by 36 to get the standard form
Now, we divide the entire equation by 36 to put it in standard form:
\[ \frac{9(x - 1)^2}{36} + \frac{4y^2}{36} = 1 \]
This simplifies to:
\[ \frac{(x - 1)^2}{4} + \frac{y^2}{9} = 1 \]

### Step 4: Identify the values of \(a\) and \(b\)
From the standard form \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\), we can identify:
- \(h = 1\) (the x-coordinate of the center)
- \(k = 0\) (the y-coordinate of the center)
- \(a^2 = 4 \Rightarrow a = 2\)
- \(b^2 = 9 \Rightarrow b = 3\)

### Step 5: Determine the vertices
Since \(b > a\), the major axis is along the y-axis. The vertices of the ellipse are given by the coordinates:
\[ (h, k \pm b) = (1, 0 \pm 3) \]
Thus, the vertices are:
\[ (1, 3) \quad \text{and} \quad (1, -3) \]

### Final Answer
The vertices of the ellipse are:
\[ (1, 3) \quad \text{and} \quad (1, -3) \]

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