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The equation given is:

\[
\frac{(x - 2)^2}{16} + \frac{(y - 5)^2}{25} = 1
\]

This is the standard form of an ellipse equation, where:

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

Here, \((h, k)\) is the center of the ellipse, \(a\) and \(b\) are the distances from the center to the vertices along the major and minor axes.

### Step-by-step solution:

1. **Identify the center**:
   From the equation, we can identify the center as \((h, k)\).
   - \(h = 2\)
   - \(k = 5\)

   So, the center is \((2, 5)\).

2. **Determine \(a^2\) and \(b^2\)**:
   - \(\frac{(x - 2)^2}{16}\) implies \(a^2 = 16\), so \(a = 4\).
   - \(\frac{(y - 5)^2}{25}\) implies \(b^2 = 25\), so \(b = 5\).

3. **Major and minor axes**:
   Since \(b = 5\) is larger than \(a = 4\), the major axis is vertical (along the \(y\)-axis). 
   
   - The major axis length is \(2b = 10\).
   - The minor axis length is \(2a = 8\).

4. **Vertices**:
   The vertices are along the major axis. Since the major axis is vertical, they are found by moving \(b\) units up and down from the center:
   - Vertex 1: \((2, 5 + 5) = (2, 10)\)
   - Vertex 2: \((2, 5 - 5) = (2, 0)\)

5. **Foci**:
   The foci are located along the major axis as well, at a distance of \(c\) from the center, where \(c\) is given by:
   \[
   c = \sqrt{b^2 - a^2} = \sqrt{25 - 16} = \sqrt{9} = 3
   \]

   The foci are then:
   - Focus 1: \((2, 5 + 3) = (2, 8)\)
   - Focus 2: \((2, 5 - 3) = (2, 2)\)

### Summary of results:
- **Center**: \((2, 5)\)
- **Vertices**: \((2, 10)\) and \((2, 0)\)
- **Foci**: \((2, 8)\) and \((2, 2)\)

Would you like more details on any of these steps?

Here are some related questions you could explore:
1. How is the general equation of an ellipse derived?
2. What are the conditions for an ellipse to become a circle?
3. How do you find the eccentricity of an ellipse?
4. Can the foci of an ellipse ever coincide with the center?
5. What happens to the vertices and foci if the ellipse is rotated?

**Tip:** When identifying the major axis, always look at the larger denominator under the squared terms.

Hallar el centro, los vértices y los focos de la elipse. Simplificar las respuestas tanto como sea posible. \[ \frac{(x - 2)^2}{16} + \frac{(y - 5)^2}{25} = 1 \]

Solve an ellipse equation by finding the center, vertices, and foci. The ellipse equation \frac{(x-2)^2}{16} + \frac{(y-5)^2}{25} = 1 is analyzed to determine key properties. This problem covers concepts in analytic geometry and is suitable for students in Grades 10-12.

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