Find the coordinates of the vertices and the foci and the length of the latus rectum of the ellipse `9x^(2)+25y^(2)=225`.

Answer

To solve the problem step by step, we will follow the standard procedure for analyzing the given ellipse equation.

### Step 1: Rewrite the equation in standard form
The given equation of the ellipse is:
\[ 9x^2 + 25y^2 = 225 \]

To convert it into standard form, we divide the entire equation by 225:
\[ \frac{9x^2}{225} + \frac{25y^2}{225} = 1 \]
This simplifies to:
\[ \frac{x^2}{25} + \frac{y^2}{9} = 1 \]

### Step 2: Identify \(a^2\) and \(b^2\)
From the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we can identify:
- \(a^2 = 25\) which gives \(a = \sqrt{25} = 5\)
- \(b^2 = 9\) which gives \(b = \sqrt{9} = 3\)

### Step 3: Find the coordinates of the vertices
For an ellipse centered at the origin with a horizontal major axis, the vertices are located at:
\[ (\pm a, 0) = (\pm 5, 0) \]
Thus, the coordinates of the vertices are:
- \( (5, 0) \)
- \( (-5, 0) \)

### Step 4: Find the coordinates of the co-vertices
The co-vertices are located at:
\[ (0, \pm b) = (0, \pm 3) \]
Thus, the coordinates of the co-vertices are:
- \( (0, 3) \)
- \( (0, -3) \)

### Step 5: Find the coordinates of the foci
The foci are located at:
\[ (\pm c, 0) \]
where \(c\) is calculated using the formula:
\[ c = \sqrt{a^2 - b^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \]
Thus, the coordinates of the foci are:
- \( (4, 0) \)
- \( (-4, 0) \)

### Step 6: Calculate the length of the latus rectum
The length of the latus rectum (LR) is given by the formula:
\[ LR = \frac{2b^2}{a} \]
Substituting the values we found:
\[ LR = \frac{2 \times 9}{5} = \frac{18}{5} \]

### Summary of Results
- **Vertices**: \( (5, 0) \) and \( (-5, 0) \)
- **Foci**: \( (4, 0) \) and \( (-4, 0) \)
- **Length of the Latus Rectum**: \( \frac{18}{5} \)