The eccentricity of the hyperbola is given by: \[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{81 / 25}{144 / 25}} = \frac{15}{12} = \frac{5}{4}. \] Hence, its focus is at \( (ae, 0) \): \[ \left(\frac{12}{5} \times \frac{5}{4}, 0 \right) = (3, 0). \] Given that the focus of the ellipse is also the same, suppose the eccentricity of the ellipse is \( e' \). Then: \[ 5e' = 3 \Rightarrow e' = \frac{3}{5}. \] Next, we have: \[ \sqrt{1 - \frac{b^2}{25}} = \frac{3}{5}. \] Squaring both sides: \[ 1 - \frac{b^2}{25} = \left(\frac{3}{5}\right)^2 = \frac{9}{25}. \] Solving for \( b^2 \): \[ \frac{b^2}{25} = 1 - \frac{9}{25} = \frac{16}{25}, \] \[ b^2 = 16. \]