To solve the problem, we need to analyze the given equations of the ellipse and hyperbola and identify their common features, specifically the center, vertices, and foci.
### Step-by-Step Solution:
1. **Identify the Equations**:
- The equation of the ellipse is given as:
\[
\frac{x^2}{25} + \frac{y^2}{16} = 1
\]
- The equation of the hyperbola is given as:
\[
\frac{x^2}{25} - \frac{y^2}{16} = 1
\]
2. **Determine the Center**:
- Both the ellipse and hyperbola are centered at the origin (0, 0). This is evident from the standard forms of the equations, where there are no additional terms added to \(x\) or \(y\).
3. **Identify the Values of \(a\) and \(b\)**:
- For the ellipse:
- \(a^2 = 25 \Rightarrow a = 5\)
- \(b^2 = 16 \Rightarrow b = 4\)
- For the hyperbola:
- \(a^2 = 25 \Rightarrow a = 5\)
- \(b^2 = 16 \Rightarrow b = 4\)
4. **Find the Vertices**:
- The vertices of the ellipse are located at:
- \((\pm a, 0) = (\pm 5, 0)\)
- The vertices of the hyperbola are also located at:
- \((\pm a, 0) = (\pm 5, 0)\)
5. **Determine the Foci**:
- For the ellipse, the distance to the foci is given by:
\[
c = \sqrt{a^2 - b^2} = \sqrt{25 - 16} = \sqrt{9} = 3
\]
- The foci of the ellipse are at \((\pm c, 0) = (\pm 3, 0)\).
- For the hyperbola, the distance to the foci is given by:
\[
c = \sqrt{a^2 + b^2} = \sqrt{25 + 16} = \sqrt{41}
\]
- The foci of the hyperbola are at \((\pm c, 0) = (\pm \sqrt{41}, 0)\).
6. **Conclusion**:
- Both the ellipse and hyperbola share the same center at the origin (0, 0).
- They also share the same vertices at \((-5, 0)\) and \((5, 0)\).
- However, they do not share the same foci, as the foci of the ellipse are at \((-3, 0)\) and \((3, 0)\), while the foci of the hyperbola are at \((-\sqrt{41}, 0)\) and \((\sqrt{41}, 0)\).
### Final Answer:
The ellipse and hyperbola have a common center and common vertices only.
