To find the vertices of the ellipse given by the equation \(9x^{2} + 25y^{2} - 90x - 150y + 225 = 0\), we will follow these steps:
### Step 1: Rearranging the Equation
First, we need to rearrange the equation into the standard form of an ellipse. We start with:
\[
9x^{2} + 25y^{2} - 90x - 150y + 225 = 0
\]
We can move the constant term to the other side:
\[
9x^{2} + 25y^{2} - 90x - 150y = -225
\]
### Step 2: Completing the Square for \(x\) and \(y\)
Next, we will complete the square for the \(x\) and \(y\) terms.
**For \(x\):**
\[
9(x^{2} - 10x)
\]
To complete the square, take half of \(-10\) (which is \(-5\)), square it (getting \(25\)), and add and subtract it inside the parentheses:
\[
9(x^{2} - 10x + 25 - 25) = 9((x - 5)^{2} - 25) = 9(x - 5)^{2} - 225
\]
**For \(y\):**
\[
25(y^{2} - 6y)
\]
Similarly, take half of \(-6\) (which is \(-3\)), square it (getting \(9\)), and add and subtract it:
\[
25(y^{2} - 6y + 9 - 9) = 25((y - 3)^{2} - 9) = 25(y - 3)^{2} - 225
\]
### Step 3: Substitute Back into the Equation
Now substituting back into the equation:
\[
9(x - 5)^{2} - 225 + 25(y - 3)^{2} - 225 = -225
\]
This simplifies to:
\[
9(x - 5)^{2} + 25(y - 3)^{2} - 450 = -225
\]
Adding \(450\) to both sides gives:
\[
9(x - 5)^{2} + 25(y - 3)^{2} = 225
\]
### Step 4: Divide by 225
To get the standard form, divide the entire equation by \(225\):
\[
\frac{(x - 5)^{2}}{25} + \frac{(y - 3)^{2}}{9} = 1
\]
### Step 5: Identify the Center and Axes
From the standard form \(\frac{(x - h)^{2}}{a^{2}} + \frac{(y - k)^{2}}{b^{2}} = 1\), we can identify:
- Center \((h, k) = (5, 3)\)
- \(a^{2} = 25 \Rightarrow a = 5\)
- \(b^{2} = 9 \Rightarrow b = 3\)
### Step 6: Find the Vertices
The vertices of the ellipse are located at:
- Along the major axis (horizontal):
- \((h - a, k) = (5 - 5, 3) = (0, 3)\)
- \((h + a, k) = (5 + 5, 3) = (10, 3)\)
Thus, the vertices of the ellipse are:
\[
(0, 3) \text{ and } (10, 3)
\]
### Final Answer
The vertices of the ellipse are \((0, 3)\) and \((10, 3)\).
---
