A hyperbola passes through the foci of the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$ and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is:
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When a hyperbola and an ellipse share the same foci, they are called confocal conics. A key property is that if a hyperbola passes through the foci of an ellipse and has the same principal axes, then the foci of the ellipse are the vertices of the hyperbola.
First, find the properties of the given ellipse: $\frac{x^2}{25} + \frac{y^2}{16} = 1$.
$a_e^2 = 25 \implies a_e = 5$.
$b_e^2 = 16 \implies b_e = 4$.
The distance from the center to the focus, $c_e$, is given by $c_e^2 = a_e^2 - b_e^2 = 25 - 16 = 9$, so $c_e = 3$.
The foci of the ellipse are at $(\pm c_e, 0) = (\pm 3, 0)$.
The eccentricity of the ellipse is $e_e = \frac{c_e}{a_e} = \frac{3}{5}$.
Now, for the hyperbola, its transverse axis is the x-axis, so its equation is of the form $\frac{x^2}{a_h^2} - \frac{y^2}{b_h^2} = 1$.
The hyperbola passes through the foci of the ellipse, $(\pm 3, 0)$. Since these points are on the transverse axis, they must be the vertices of the hyperbola.
Therefore, the vertex of the hyperbola is at $(\pm a_h, 0) = (\pm 3, 0)$, which gives $a_h = 3$.
We are given that the product of their eccentricities is 1: $e_e \cdot e_h = 1$.
$\frac{3}{5} \cdot e_h = 1 \implies e_h = \frac{5}{3}$.
For a hyperbola, the eccentricity is related to its semi-axes by $e_h^2 = 1 + \frac{b_h^2}{a_h^2}$.
Substituting the values we found:
$\left(\frac{5}{3}\right)^2 = 1 + \frac{b_h^2}{3^2}$.
$\frac{25}{9} = 1 + \frac{b_h^2}{9}$.
$\frac{b_h^2}{9} = \frac{25}{9} - 1 = \frac{16}{9}$.
$b_h^2 = 16$.
The equation of the hyperbola is $\frac{x^2}{a_h^2} - \frac{y^2}{b_h^2} = 1$, which is $\frac{x^2}{9} - \frac{y^2}{16} = 1$.
