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Step-by-Step Solution
Step 1: Identify the Equation of the Ellipse
The given ellipse is
$ \frac{x^2}{25} + \frac{y^2}{b^2} = 1 $
with the condition $b < 5$. Let its eccentricity be $e_1$.
For an ellipse in standard form
$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $,
if $a^2 = 25$, then $a = 5$. The relationship between $a, b, e$
for an ellipse is
$ b^2 = a^2 (1 - e^2). $
Step 2: Write the Relation for the Ellipse
From the ellipse equation,
$ b^2 = 25(1 - e_1^2). \quad (1) $
Step 3: Identify the Equation of the Hyperbola
The given hyperbola is
$ \frac{x^2}{16} - \frac{y^2}{b^2} = 1. $
Let its eccentricity be $e_2$. For a hyperbola of the form
$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $,
if $a^2 = 16$, then $a = 4$. The relation for a hyperbola is
$ b^2 = a^2(e^2 - 1). $
Step 4: Write the Relation for the Hyperbola
From the hyperbola equation,
$ b^2 = 16(e_2^2 - 1). \quad (2) $
Step 5: Use the Given Condition on Eccentricities
We are given that
$ e_1 \cdot e_2 = 1. $
Hence, $ e_2 = \frac{1}{e_1}. $
Step 6: Equate the Two Expressions for $b^2$
From (1) and (2), the same $b^2$ must satisfy:
$ 25(1 - e_1^2) = 16(e_2^2 - 1). $
But $e_2 = \frac{1}{e_1}$, so $e_2^2 = \frac{1}{e_1^2}.$ Substitute into the equation:
$ 25(1 - e_1^2) = 16\left(\frac{1}{e_1^2} - 1\right). $
Step 7: Solve for $e_1$
Rewriting gives:
$ 25(1 - e_1^2) = 16\left(\frac{1 - e_1^2}{e_1^2}\right). $
Simplify the right-hand side:
$ 25(1 - e_1^2) = \frac{16(1 - e_1^2)}{e_1^2}. $
Assuming $(1 - e_1^2) \neq 0$, we can divide both sides by $(1 - e_1^2)$:
$ 25 = \frac{16}{e_1^2} \quad \Longrightarrow \quad e_1^2 = \frac{16}{25} \quad \Longrightarrow \quad e_1 = \frac{4}{5}. $
Step 8: Determine $e_2$
Since $e_1 \cdot e_2 = 1$, we get
$ e_2 = \frac{1}{e_1} = \frac{5}{4}. $
Step 9: Calculate the Distance Between the Foci of the Ellipse
For an ellipse, the distance between its two foci is $2 a e$. Here $a = 5$ and $e_1 = \frac{4}{5}$,
so
$ \alpha = 2 \times 5 \times \frac{4}{5} = 8. $
Step 10: Calculate the Distance Between the Foci of the Hyperbola
For a hyperbola, the distance between its two foci is $2 a e$. Here $a = 4$ and $e_2 = \frac{5}{4}$,
so
$ \beta = 2 \times 4 \times \frac{5}{4} = 10. $
Final Answer
Thus, the ordered pair $(\alpha, \beta)$ is $\boxed{(8,\,10)}$.
