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Step-by-Step Solution
Step 1: Identify the ellipse and its eccentricity
The given ellipse is
$ \frac{x^2}{25} + \frac{y^2}{16} = 1.$
For an ellipse of the form $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ with $a^2 = 25$ and $b^2 = 16,$ the eccentricity $e_1$ is given by
$$
e_1 = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}.
$$
Step 2: Determine the foci of the ellipse
Since $a = 5$ and $e_1 = \frac{3}{5},$ the foci of the ellipse are at
$ (\pm a e_1,\, 0) \;=\; (\pm 3,\, 0).$
Step 3: Express the general form of the hyperbola
We are told the hyperbola's transverse axis (along x-axis) and conjugate axis (along y-axis) coincide with those of the ellipse. Hence, we assume the hyperbola has the form
$$
\frac{x^2}{A^2} \;-\; \frac{y^2}{B^2} \;=\; 1.
$$
Since it passes through the ellipse's foci $(\pm3,\,0),$ plugging in $(3,0)$ gives
$$
\frac{3^2}{A^2} = 1 \quad \Longrightarrow \quad \frac{9}{A^2} = 1 \quad \Longrightarrow \quad A^2 = 9.
$$
Thus, $A = 3.$
Step 4: Use the condition on product of eccentricities
We are given that the product of the eccentricity of the ellipse ($e_1$) and that of the hyperbola ($e_2$) is 1, i.e.
$$
e_1 \cdot e_2 = 1.
$$
Since $e_1 = \frac{3}{5},$ it follows that
$$
\left(\frac{3}{5}\right)\, e_2 = 1
\quad \Longrightarrow \quad
e_2 = \frac{5}{3}.
$$
Step 5: Find $B^2$ using the hyperbola’s eccentricity
For a hyperbola of the form
$ \frac{x^2}{A^2} - \frac{y^2}{B^2} = 1, $
the eccentricity $e_2$ is given by
$$
e_2 = \sqrt{\,1 + \frac{B^2}{A^2}}.
$$
Substituting $e_2 = \frac{5}{3}$ and $A^2 = 9,$ we get
$$
\left(\frac{5}{3}\right)^2 = 1 + \frac{B^2}{9}.
$$
Therefore,
$$
\frac{25}{9} = 1 + \frac{B^2}{9}
\quad \Longrightarrow \quad
\frac{25}{9} - 1 = \frac{B^2}{9}
\quad \Longrightarrow \quad
\frac{16}{9} = \frac{B^2}{9}
\quad \Longrightarrow \quad
B^2 = 16.
$$
Step 6: Write the equation of the hyperbola
With $A^2 = 9$ and $B^2 = 16,$ the hyperbola equation becomes
$$
\frac{x^2}{9} - \frac{y^2}{16} = 1.
$$
