Here is a more elaborate, clear, and educational solution to the problem.

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1. Key Concepts and Formulas

To effectively solve this problem, we need a solid understanding of the standard forms and properties of ellipses and hyperbolas centered at the origin. It's crucial to use distinct notations for the parameters of the ellipse and the hyperbola to avoid confusion.

• Ellipse (Major axis along the x-axis): The standard equation is: $\frac{x^2}{a_e^2} + \frac{y^2}{b_e^2} = 1 \quad \text{where } a_e > b_e$

  • Here, $a_e$ is the semi-major axis length and $b_e$ is the semi-minor axis length.
  • Its eccentricity, denoted by $e_e$, is given by: $e_e = \sqrt{1 - \frac{b_e^2}{a_e^2}} \quad \text{or equivalently, } b_e^2 = a_e^2(1 - e_e^2)$
  • Its foci are located at: $(\pm a_e e_e, 0)$

• Hyperbola (Transverse axis along the x-axis): The standard equation is: $\frac{x^2}{a_h^2} - \frac{y^2}{b_h^2} = 1$

  • Here, $a_h$ is the semi-transverse axis length and $b_h$ is the semi-conjugate axis length.
  • Its eccentricity, denoted by $e_h$, is given by: $e_h = \sqrt{1 + \frac{b_h^2}{a_h^2}} \quad \text{or equivalently, } b_h^2 = a_h^2(e_h^2 - 1)$
  • Its vertices are located at: $(\pm a_h, 0)$
  • Its foci are located at: $(\pm a_h e_h, 0)$

2. Step-by-Step Solution

Let's break down the problem into manageable steps, extracting information from the ellipse first, then applying it to the hyperbola.

Step 1: Analyze the given Ellipse

We are given the equation of the ellipse: $\frac{x^2}{25} + \frac{y^2}{16} = 1$

• Identify $a_e^2$ and $b_e^2$: By comparing this with the standard form $\frac{x^2}{a_e^2} + \frac{y^2}{b_e^2} = 1$,