This problem requires a comprehensive understanding of the properties of both hyperbolas and ellipses, and the ability to connect them through given conditions. We will systematically determine the parameters of each conic section and then use them to find the required chord length.

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1. Key Concepts and Formulas

Before we dive into the solution, let's recall the fundamental definitions and formulas for hyperbolas and ellipses centered at the origin:

• Hyperbola ($\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$, transverse axis along $x$-axis):
  • Eccentricity ($e_H$): $e_H = \sqrt{1 + \frac{B^2}{A^2}}$
  • Foci: $(\pm A e_H, 0)$
• Ellipse ($\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, major axis along $x$-axis, $a > b$):
  • Eccentricity ($e_E$): $e_E = \sqrt{1 - \frac{b^2}{a^2}}$
  • Relationship between $a, b, e_E$: $b^2 = a^2(1 - e_E^2)$
• Chord of an Ellipse: A horizontal chord passing through $y=k$ intersects the ellipse at two points $(x_1, k)$ and $(x_2, k)$. Its length is $|x_2 - x_1|$. Due to symmetry, if the ellipse is centered at the origin, the points will be $(\pm x, k)$, and the length will be $2|x|$.

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2. Step-by-Step Solution

Step 1: Determine the Eccentricity and Foci of the Given Hyperbola

We are given the equation of the hyperbola: $\frac{x^2}{16} - \frac{y^2}{9} = 1$

Concept: The standard form of a hyperbola with its transverse axis along the $x$-axis is $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$. Its eccentricity is $e_H = \sqrt{1 + \frac{B^2}{A^2}}$ and its foci are at $(\pm A e_H, 0)$.

Working:

1. By comparing the given equation with the standard form, we identify: $A^2 = 16 \implies A = 4$ $B^2 = 9 \implies B = 3$
2. Now, calculate the eccentricity $e_1$ of the hyperbola: $e_1 = \sqrt{1 + \frac{B^2}{A^2}} = \sqrt{1 + \frac{9}{16}}$ $e_1 = \sqrt{\frac{16+9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$
3. Next, find the coordinates of the foci of the hyperbola: $\text{Foci} = (\pm A e_1, 0) = \left(\pm 4 \cdot \frac{5}{4}, 0\right) = (\pm 5, 0)$

Explanation: We extract the fundamental parameters $A$ and $B$ from the hyperbola's equation. These parameters are essential for calculating its eccentricity $e_1$, which quantifies the shape of the hyperbola, and its foci, which are key defining points. These foci will then be used as points through which the ellipse passes.

Tip: It's crucial to correctly identify $A^2$ and $B^2$. For a hyperbola $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$, $A^2$ is always under the $x^2$ term and $B^2$ under the $y^2$ term when the transverse axis is along the $x$-axis.

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Step 2: Determine the Major Semi-axis ($a$) of the Ellipse

The problem states that the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ passes through the foci of the hyperbola, which we found to be $(\pm 5, 0)$.

Concept: If a point lies on a curve, its coordinates must satisfy the equation of the curve.

Working:

1. Substitute the coordinates of one of the foci, say $(5, 0)$, into the ellipse's equation: $\frac{5^2}{a^2} + \frac{0^2}{b^2} = 1$ $\frac{25}{a^2} + 0 = 1$ $\frac{25}{a^2} = 1 \implies a^2 = 25$

Explanation: By substituting the coordinates of the hyperbola's foci into the ellipse's equation, we establish a direct relationship that allows us to determine the value of $a^2$ for the ellipse. This is a critical step towards defining the specific ellipse. Since $a^2=25 \implies a=5$, the points $(\pm 5, 0)$ are actually the vertices of the ellipse.

Common Mistake: Do not assume that the foci of the hyperbola are also the foci of the ellipse. The problem states the ellipse passes through these points, which is a distinct condition. In this case, they happen to be the vertices of the ellipse.

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Step 3: Determine the Eccentricity ($e_2$) and $b^2$ of the Ellipse

We are given the condition $e_1 e_2 = 1$.

Concept: We use the given relationship between the eccentricities to find $e_2$, and then use the standard formula for ellipse eccentricity, $e_E = \sqrt{1 - \frac{b^2}{a^2}}$, to relate it to $a^2$ and $b^2$.

Working:

1. Using the given condition and $e_1 = \frac{5}{4}$ (from Step 1): $\frac{5}{4} \cdot e_2 = 1 \implies e_2 = \frac{4}{5}$
2. Now, use the eccentricity formula for an ellipse: $e_2 = \sqrt{1 - \frac{b^2}{a^2}}$ $\frac{4}{5} = \sqrt{1 - \frac{b^2}{a^2}}$
3. Square both sides to remove the square root: $\left(\frac{4}{5}\right)^2 = 1 - \frac{b^2}{a^2}$ $\frac{16}{25} = 1 - \frac{b^2}{a^2}$
4. Rearrange to solve for $\frac{b^2}{a^2}$: $\frac{b^2}{a^2} = 1 - \frac{16}{25} = \frac{25 - 16}{25} = \frac{9}{25}$

Explanation: The product of eccentricities provides a direct link between the two conic sections. Once $e_2$ is found, we use its definition to establish a relationship between $a^2$ and $b^2$, which are the remaining unknown parameters for the ellipse.

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Step 4: Formulate the Complete Equation of the Ellipse

Concept: With $a^2$ determined from the points the ellipse passes through (Step 2) and the ratio $\frac{b^2}{a^2}$ determined from its eccentricity (Step 3), we can now find $b^2$ and write the full equation of the ellipse.

Working:

1. From Step 2, we have $a^2 = 25$.
2. From Step 3, we have $\frac{b^2}{a^2} = \frac{9}{25}$.
3. Substitute $a^2 = 25$ into the ratio: $\frac{b^2}{25} = \frac{9}{25} \implies b^2 = 9$
4. Therefore, the equation of the ellipse is: $\frac{x^2}{25} + \frac{y^2}{9} = 1$

Explanation: This step completes the identification of the specific ellipse by finding both $a^2$ and $b^2$. Having the full equation is essential for the final calculation of the chord length.

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Step 5: Calculate the Length of the Chord of the Ellipse

We need to find the length of the chord of the ellipse that is parallel to the $x$-axis and passes through the point $(0,2)$.

Concept: A chord parallel to the $x$-axis means that all points on the chord have the same $y$-coordinate. If the chord passes through $(0,2)$, then its equation is $y=2$. To find the length, we substitute $y=2$ into the ellipse equation to find the corresponding $x$-coordinates of the intersection points. The length is then $2|x|$.

Working:

1. Substitute $y=2$ into the ellipse equation $\frac{x^2}{25} + \frac{y^2}{9} = 1$: $\frac{x^2}{25} + \frac{2^2}{9} = 1$ $\frac{x^2}{25} + \frac{4}{9} = 1$
2. Solve for $x^2$: $\frac{x^2}{25} = 1 - \frac{4}{9}$ $\frac{x^2}{25} = \frac{9 - 4}{9} = \frac{5}{9}$ $x^2 = 25 \cdot \frac{5}{9} = \frac{125}{9}$
3. Take the square root to find $x$: $x = \pm \sqrt{\frac{125}{9}} = \pm \frac{\sqrt{25 \cdot 5}}{\sqrt{9}} = \pm \frac{5\sqrt{5}}{3}$
4. The two points where the chord intersects the ellipse are $\left(-\frac{5\sqrt{5}}{3}, 2\right)$ and $\left(\frac{5\sqrt{5}}{3}, 2\right)$.
5. The length of the chord is the distance between these two points. Since the $y$-coordinates are the same, it's simply the difference in $x$-coordinates: $\text{Length of chord} = \left|\frac{5\sqrt{5}}{3} - \left(-\frac{5\sqrt{5}}{3}\right)\right| = \left|\frac{5\sqrt{5}}{3} + \frac{5\sqrt{5}}{3}\right|$ $\text{Length of chord} = 2 \cdot \frac{5\sqrt{5}}{3} = \frac{10\sqrt{5}}{3}$

Explanation: This final step applies the specific equation of the ellipse we derived. By setting the $y$-coordinate to $2$, we find the $x$-coordinates of the two points where the horizontal line $y=2$ intersects the ellipse. The distance between these two $x$-coordinates gives the required chord length.

Tip: For a chord parallel to the $x$-axis, the length is always $2|x|$ where $x$ is the positive $x$-coordinate of the intersection point. This avoids potential sign errors when calculating the difference.

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3. Final Answer Check

• Hyperbola: $\frac{x^2}{16} - \frac{y^2}{9} = 1 \implies A=4, B=3$.
• $e_1 = \sqrt{1 + 9/16} = \sqrt{25/16} = 5/4$.
• Hyperbola foci: $(\pm A e_1, 0) = (\pm 4 \cdot