1. Introduction and Key Concepts

This problem is a comprehensive test of your understanding of conic sections, specifically ellipses and hyperbolas. We need to leverage their fundamental properties: the definition of foci, eccentricity, and the unique property of focal distances for a hyperbola.

Let's review the essential formulas and definitions that will guide our solution:

• Ellipse: For a standard ellipse centered at the origin, with equation $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$:

  • Orientation:
    • If $B > A$, it's a vertical ellipse. The major axis is along the y-axis, with semi-major axis length $B$ and semi-minor axis length $A$.
    • If $A > B$, it's a horizontal ellipse. The major axis is along the x-axis, with semi-major axis length $A$ and semi-minor axis length $B$.
  • Foci: The distance from the center to each focus is denoted by $c_E$. The relationship is $c_E^2 = (\text{semi-major axis})^2 - (\text{semi-minor axis})^2$.
    • For a vertical ellipse: $c_E^2 = B^2 - A^2$. Foci are at $(0, \pm c_E)$.
    • For a horizontal ellipse: $c_E^2 = A^2 - B^2$. Foci are at $(\pm c_E, 0)$.
  • Eccentricity ($e_E$): $e_E = \frac{c_E}{\text{semi-major axis}}$.
    • For a vertical ellipse: $e_E = \frac{c_E}{B}$.
    • For a horizontal ellipse: $e_E = \frac{c_E}{A}$.

• Hyperbola: For a standard hyperbola centered at the origin:

  • Orientation:
    • If its foci are on the y-axis at $(0, \pm c_H)$, it's a vertical hyperbola. Its standard equation is $\frac{y^2}{a_H^2} - \frac{x^2}{b_H^2} = 1$, where $a_H$ is the length of the semi-transverse axis.
    • If its foci are on the x-axis at $(\pm c_H, 0)$, it's a horizontal hyperbola. Its standard equation is $\frac{x^2}{a_H^2} - \frac{y^2}{b_H^2} = 1$, where $a_H$ is the length of the semi-transverse axis.
  • Foci: The distance from the center to each focus is $c_H$. The relationship between $a_H$, $b_H$, and $c_H$ is $c_H^2 = a_H^2 + b_H^2$.
  • Eccentricity ($e_H$): $e_H = \frac{c_H}{a_H}$.
  • Focal Distance Property: For any point $P(x,y)$ on a hyperbola, the distances from $P$ to the two foci $S_1$ and $S_2$ (called focal distances) are given by specific formulas.
    • For a horizontal hyperbola (foci on x-axis): $PS_1 = |e_H x - a_H|$ and $PS_2 = |e_H x + a_H|$.
    • For a vertical hyperbola (foci on y-axis): $PS_1 = |e_H y - a_H|$ and $PS_2 = |e_H y + a_H|$.
    • A defining property of a hyperbola is that the absolute difference of these focal distances is constant and equal to $2a_H$, i.e., $|PS_1 - PS_2| = 2a_H$.

Our strategy will be to first extract all relevant information from the given ellipse, then use that to define the hyperbola's key parameters, and finally apply the focal distance formula to the given point.

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2. Step-by-Step Solution

Step 1: Analyze the Given Ellipse

We are given the equation of the ellipse: $\frac{x^2}{9} + \frac{y^2}{25} = 1$

• Identify Semi-axes lengths: Comparing this to the standard form $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$, we can identify the squares of the semi-axes: $A^2 = 9 \implies A = 3$ $B^2 = 25 \implies B = 5$

• Determine Orientation: Since $B > A$ ($5 > 3$), the denominator under $y^2$ is larger. This indicates that the major axis lies along the y-axis, making it a vertical ellipse.

  • Why this is important: The orientation determines the formula for $c_E$ and $e_E$, and where the foci are located.

• Calculate Focal Distance ($c_E$) for the Ellipse: For a vertical ellipse, the relationship between $A, B,$ and $c_E$ is $c_E^2 = B^2 - A^2$. $c_E^2 = 25 - 9 = 16$ $c_E = \sqrt{16} = 4$

• Determine Foci of the Ellipse: The foci of a vertical ellipse are located at $(0, \pm c_E)$. Thus, the foci of the ellipse are $(0, \pm 4)$.

• Calculate Eccentricity of the Ellipse ($e_E$): For a vertical ellipse, the eccentricity is given by $e_E = \frac{c_E}{B}$ (where $B$ is the semi-major axis length). $e_E = \frac{4}{5}$

Tip: Always double-check the values of $A^2$ and $B^2$ to correctly identify the major axis and thus the orientation of the ellipse. This is a common source of error.

Step 2: Determine the Hyperbola's Foci and Eccentricity

The problem states that the foci of the hyperbola are the same as that of the ellipse.

• Foci of the Hyperbola: From Step 1, the foci of the ellipse are $(0, \pm 4)$. Therefore, the foci of the hyperbola are also $(0, \pm 4)$.

  • Why this is crucial:
    1. Since the foci lie on the y-axis, the hyperbola must be a vertical hyperbola. This dictates the form of its standard equation and the focal distance formulas we'll use.
    2. The distance from the center (origin) to each focus for the hyperbola is $c_H = 4$.

• Eccentricity of the Hyperbola ($e_H$): We are given a relationship between the eccentricities: the eccentricity of the hyperbola is $\frac{15}{8}$ times the eccentricity of the ellipse. $e_H = \frac{15}{8} \times e_E$ Substitute the value of $e_E = \frac{4}{5}$ (calculated in Step 1): $e_H = \frac{15}{8} \times \frac{4}{5}$ $e_H = \frac{3 \times 5}{2 \times 4} \times \frac{4}{5}$ $e_H = \frac{3}{2}$ So, the eccentricity of the hyperbola is $e_H = \frac{3}{2}$.

Step 3: Find the Semi-Transverse Axis ($a_H$) of the Hyperbola

For any hyperbola, the relationship between its focal distance $c_H$, semi-transverse axis $a_H$, and eccentricity $e_H$ is given by $c_H = a_H e_H$.

• We know $c_H = 4$ (from Step 2) and $e_H = \frac{3}{2}$ (from Step 2). We can now solve for $a_H$. $4 = a_H \times \frac{3}{2}$ To isolate $a_H$, multiply both sides by $\frac{2}{3}$: $a_H = 4 \times \frac{2}{3} = \frac{8}{3}$ So, the length of the semi-transverse axis of the hyperbola is $a_H = \frac{8}{3}$.
  • Why we need $a_H$: This value is essential for calculating the focal distances using the property $PS = |e_H y \pm a_H|$.

Step 4: Calculate the Focal Distances for the given Point

We need to find the smaller focal distance of the point $P\left(\sqrt{2}, \frac{14}{3} \sqrt{\frac{2}{5}}\right)$ on the hyperbola. Let the coordinates of the point be $(x_P, y_P)$, so $x_P = \sqrt{2}$ and $y_P = \frac{14}{3} \sqrt{\frac{2}{5}}$.

Since it's a vertical hyperbola (foci on y-axis), the focal distances for a point $P(x_P,y_P)$ are: $PS_1 = |e_H y_P - a_H|$ $PS_2 = |e_H y_P + a_H|$

First, let's calculate the product $e_H y_P$:

• $y_P = \frac{14}{3} \sqrt{\frac{2}{5}}$
• $e_H = \frac{3}{2}$ $e_H y_P = \frac{3}{2} \times \frac{14}{3} \sqrt{\frac{2}{5}}$ $e_H y_P = \left(\frac{3 \times 14}{2 \times 3}\right) \sqrt{\frac{2}{5}}$ $e_H y_P = 7 \sqrt{\frac{2}{5}}$

Now, substitute $e_H y_P = 7 \sqrt{\frac{2}{5}}$ and $a_H = \frac{8}{3}$ into the focal distance formulas:

• $PS_1 = \left| 7 \sqrt{\frac{2}{5}} - \frac{8}{3} \right|$
• $PS_2 = \left| 7 \sqrt{\frac{2}{5}} + \frac{8}{3} \right|$

Let's evaluate the terms:

• For $PS_2$: Since both $7 \sqrt{\frac{2}{5}}$ and $\frac{8}{3}$ are positive values, their sum is positive. $PS_2 = 7 \sqrt{\frac{2}{5}} + \frac{8}{3}$

• For $PS_1$: We need to determine if $7 \sqrt{\frac{2}{5}} - \frac{8}{3}$ is positive or negative to correctly remove the absolute value sign. Let's compare $7 \sqrt{\frac{2}{5}}$ and $\frac{8}{3}$: To compare them easily, we can square both values: $(7 \sqrt{\frac{2}{5}})^2 = 49 \times \frac{2}{5} = \frac{98}{5} = 19.6$ $(\frac{8}{3})^2 = \frac{64}{9} \approx 7.11$ Since $19.6 > 7.11$, it means $7 \sqrt{\frac{2}{5}} > \frac{8}{3}$. Therefore, the term $7 \sqrt{\frac{2}{5}} - \frac{8}{3}$ is positive. So, $PS_1 = 7 \sqrt{\frac{2}{5}} - \frac{8}{3}$.

• Identify the Smaller Focal Distance: Comparing $PS_1 = 7 \sqrt{\frac{2}{5}} - \frac{8}{3}$ and $PS_2 = 7 \sqrt{\frac{2}{5}} + \frac{8}{3}$, it is evident that $PS_1$ is the smaller focal distance because a positive value is being subtracted from $7 \sqrt{\frac{2}{5}}$ in $PS_1$, while it's being added in $PS_2$.

The smaller focal distance is $7 \sqrt{\frac{2}{5}} - \frac{8}{3}$.

Common Mistake: A frequent error is forgetting to take the absolute value for focal distances, or incorrectly determining the sign of the expression inside the absolute value. Always explicitly compare $e_H y_P$ and $a_H$ (or $e_H x_P$ and $a_H$) to ensure the correct sign is applied.

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3. Summary and Key Takeaway

In this problem, we systematically used the definitions and properties of conic sections.

1. We identified the ellipse's orientation and calculated its foci and eccentricity.
2. We then used the shared foci to determine the hyperbola's orientation and its focal distance $c_H$. The given relationship allowed us to find the hyperbola's eccentricity $e_H$.
3. With $c_H$ and $e_H$, we found the hyperbola's semi-transverse axis $a_H$.
4. Finally, we applied the specific focal distance formulas for a vertical hyperbola to the given point, carefully handling the absolute value to find the smaller distance.

This problem highlights the interconnectedness of conic section properties and emphasizes the importance of correctly identifying the orientation