This problem is a comprehensive test of your understanding of conic sections, specifically hyperbolas and ellipses. It requires you to correctly identify parameters from their standard equations, apply formulas for vertices, eccentricities, and the latus rectum, and interpret conditions that relate these properties between the two curves. We will break down the problem step-by-step, explaining the rationale behind each action.

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1. Key Concepts and Formulas for Conic Sections

Before we begin, let's establish the standard forms and associated properties for the hyperbola and ellipse, paying close attention to their orientation.

• Hyperbola ($H$):

  • Standard Form (Transverse axis along y-axis): $\frac{y^2}{A^2} - \frac{x^2}{B^2} = 1$
    • This form indicates that the positive term is associated with $y^2$, meaning the transverse axis (the axis containing the foci and vertices) lies along the y-axis. The length of the semi-transverse axis is $A$, and the length of the semi-conjugate axis is $B$.
  • Vertices: $(0, \pm A)$
  • Eccentricity ($e_H$): $e_H = \sqrt{1 + \frac{B^2}{A^2}}$
    • Eccentricity measures how "open" the hyperbola is. For a hyperbola, $e_H > 1$.

• Ellipse ($E$):

  • Standard Form (Major axis along y-axis): $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
    • This form implies that $b > a$, meaning the major axis (the longer axis containing the foci and vertices) lies along the y-axis. The semi-major axis is $b$, and the semi-minor axis is $a$.
  • Vertices (endpoints of major axis): $(0, \pm b)$
  • Vertices (endpoints of minor axis): $(\pm a, 0)$
  • Eccentricity ($e_E$): $e_E = \sqrt{1 - \frac{a^2}{b^2}}$ (since $b>a$)
    • Eccentricity measures how "flat" the ellipse is. For an ellipse, $0 < e_E < 1$.
  • Length of Latus Rectum ($l$): $l = \frac{2a^2}{b}$ (when the major axis is along the y-axis)
    • The latus rectum is a chord passing through a focus and perpendicular to the major axis. Its length is a key characteristic of the ellipse.

Tip: Always begin by identifying the orientation of the major/transverse axis. This is crucial as it determines which parameter ($a$ or $b$ for ellipse, $A$ or $B$ for hyperbola) corresponds to the semi-axis length along a particular coordinate axis and dictates the correct formulas for eccentricity and latus rectum.

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2. Analyzing the Hyperbola ($H$)

Our first step is to extract all necessary information from the given equation of the hyperbola, $H$. We need its standard form, the lengths of its semi-axes, its vertices, and its eccentricity.

Given equation: $H: \frac{x^2}{49} - \frac{y^2}{64} = -1$

2.1.1. Rewrite in Standard Form and Identify Parameters

• Why this step? The standard form of a hyperbola has a positive term first. Rewriting the equation in this manner allows for direct comparison with the general standard forms, helping us correctly identify the orientation of the transverse axis and the values of $A^2$ and $B^2$.
• We multiply the entire equation by $-1$ to make the $y^2$ term positive: $-\left(\frac{x^2}{49}\right) + \left(\frac{y^2}{64}\right) = 1$ $\frac{y^2}{64} - \frac{x^2}{49} = 1$
• Key Concept: Compare this with the standard form of a hyperbola with its transverse axis along the y-axis: $\frac{y^2}{A^2} - \frac{x^2}{B^2} = 1$
• From this comparison, we identify the values of $A^2$ and $B^2$:
  • $A^2 = 64 \implies A = 8$ (since $A$ represents a length, it must be positive). $A$ is the length of the semi-transverse axis.
  • $B^2 = 49 \implies B = 7$ (since $B$ represents a length, it must be positive). $B$ is the length of the semi-conjugate axis.

2.1.2. Determine Transverse Axis and Vertices

• Why this step? The problem states that the ellipse passes through the vertices of the hyperbola. Correctly identifying these vertices is crucial for finding the ellipse's parameters.
• Key Concept: Since the $y^2$ term is positive in the standard form $\frac{y^2}{A^2} - \frac{x^2}{B^2} = 1$, the transverse axis of the hyperbola is along the y-axis.
• The vertices of a hyperbola with its transverse axis along the y-axis are given by $(0, \pm A)$.
• Substituting $A=8$, the vertices of hyperbola $H$ are $(0, \pm 8)$.

2.1.3. Calculate Eccentricity of Hyperbola ($e_H$)

• Why this step? The problem provides a condition involving the product of the eccentricities of the ellipse and hyperbola. We need $e_H$ to use this condition.
• Key Concept: The eccentricity of a hyperbola is given by the formula: $e_H = \sqrt{1 + \frac{B^2}{A^2}}$
• Substitute the values $A^2=64$ and $B^2=49$: $e_H = \sqrt{1 + \frac{49}{64}}$ $e_H = \sqrt{\frac{64+49}{64}}$ $e_H = \sqrt{\frac{113}{64}}$ $e_H = \frac{\sqrt{113}}{8}$
• Tip: Always check that the eccentricity of a hyperbola is greater than 1. Here, $\frac{\sqrt{113}}{8} \approx \frac{10.6}{8} > 1$, which is consistent.

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3. Analyzing the Ellipse ($E$) - Part 1: Axis Orientation and Major Axis Length

Now we turn our attention to the ellipse $E$, using the information derived from the hyperbola and the conditions given in the problem.

Given conditions for Ellipse ($E$):

1. It passes through the vertices of the hyperbola $H$.
2. Its major and minor axes coincide with the transverse and conjugate axes of the hyperbola $H$, respectively.

3.1.1. Determine Axis Orientation of Ellipse ($E$)

• Why this step? The orientation of the major axis dictates the correct standard equation for the ellipse and the formulas for its eccentricity and latus rectum.
• From Step 2.1.2, we know the hyperbola $H$ has its transverse axis along the y-axis.
• The problem states that the ellipse's major axis coincides with the hyperbola's transverse axis.
• Therefore, the major axis of the ellipse $E$ is along the y-axis.
• Key Concept: For an ellipse with its major axis along the y-axis, the semi-major axis is denoted by $b$ (along the y-axis) and the semi-minor axis by $a$ (along the x-axis), implying $b > a$.
• The standard equation for such an ellipse is: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

3.1.2. Use Vertices Information to Find $b$

• Why this step? The fact that the ellipse passes through specific points allows us to determine one of its key parameters, $b$.
• From Step 2.1.2, the vertices of the hyperbola are $(0, \pm 8)$. The ellipse $E$ passes through these points.
• Since the major axis of the ellipse is along the y-axis, its vertices (endpoints of the major axis) are $(0, \pm b)$.
• Therefore, the points $(0, \pm 8)$ are the vertices of the ellipse $E$.
• Comparing $(0, \pm b)$ with $(0, \pm 8)$, we directly find: $b = 8$
• Squaring this value, we get: $b^2 = 64$

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4. Analyzing the Ellipse ($E$) - Part 2: Using Eccentricity Product to Find $a$

Now we have $b$ for the ellipse. To find the length of the latus rectum, we also need $a$. We will use the given condition about the product of eccentricities.

Given condition: The product of the eccentricities of $E$ and $H$ is $\frac{1}{2}$, i.e., $e_E \cdot e_H = \frac{1}{2}$

4.1.1. Express Eccentricity of Ellipse ($e_E$) in terms of $a$

• Why this step? We need an expression for $e_E$ that involves the unknown $a$ so we can solve for $a$ using the product condition.
• Key Concept: For an ellipse with its major axis along the y-axis ($b>a$), the eccentricity formula is: $e_E = \sqrt{1 - \frac{a^2}{b^2}}$
• Substitute the value $b^2 = 64$ that we found in Step 3.1.2: $e_E = \sqrt{1 - \frac{a^2}{64}}$ $e_E = \sqrt{\frac{64 - a^2}{64}}$ $e_E = \frac{\sqrt{64 - a^2}}{8}$
• Tip: Always check that the eccentricity of an ellipse is between 0 and 1. This expression will yield a value between 0 and 1 as long as $0 < a^2 < 64$.

4.1.2. Solve for $a^2$ using the Product of Eccentricities

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