Passage: The foci of the ellipse px^2+16y^2=16p and the foci of the...

The equation of the ellipse can be written as $\dfrac{x^2}{16} + \dfrac{y^2}{p} = 1$.

Since $p < 16$, the major axis is along the x-axis. The eccentricity $e_1$ of the ellipse is given by $e_1 = \sqrt{1 - \dfrac{p}{16}}$.

The foci of the ellipse are $(\pm 4e_1, 0) = \left(\pm 4\sqrt{1 - \dfrac{p}{16}}, 0\right) = (\pm \sqrt{16 - p}, 0)$.

The equation of the hyperbola is $25(81x^2 - 144y^2) = 11664$.

Dividing by $11664$, we get $\dfrac{25x^2}{144} - \dfrac{25y^2}{81} = 1 \Rightarrow \dfrac{x^2}{144/25} - \dfrac{y^2}{81/25} = 1$.

The eccentricity $e_2$ of the hyperbola is $e_2 = \sqrt{1 + \dfrac{81/25}{144/25}} = \sqrt{1 + \dfrac{81}{144}} = \sqrt{\dfrac{225}{144}} = \dfrac{15}{12} = \dfrac{5}{4} = 1.25$.

The foci of the hyperbola are $(\pm A e_2, 0) = \left(\pm \dfrac{12}{5} \times \dfrac{5}{4}, 0\right) = (\pm 3, 0)$.

Since the foci of the ellipse and the hyperbola coincide, we have $\sqrt{16 - p} = 3 \Rightarrow 16 - p = 9 \Rightarrow p = 7$.

The eccentricity of the ellipse is $e_1 = \sqrt{1 - \dfrac{7}{16}} = \sqrt{\dfrac{9}{16}} = \dfrac{3}{4} = 0.75$.

The difference between the eccentricities is $e_2 - e_1 = 1.25 - 0.75 = 0.5$.

Answer: $0\cdot5$