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

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

Since $p < 16$, the major axis is along the x-axis. The foci are given by $(\pm \sqrt{a^2 - b^2}, 0) = (\pm \sqrt{16 - p}, 0)$.

The equation of the hyperbola is $25(81x^2 - 144y^2) = 11664$, which can be rewritten as $\dfrac{x^2}{144/25} - \dfrac{y^2}{81/25} = 1$.

For the hyperbola, $a^2 = \dfrac{144}{25}$ and $b^2 = \dfrac{81}{25}$. The foci are given by $(\pm \sqrt{a^2 + b^2}, 0) = \left(\pm \sqrt{\dfrac{144}{25} + \dfrac{81}{25}}, 0\right) = \left(\pm \sqrt{\dfrac{225}{25}}, 0\right) = (\pm 3, 0)$.

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

Squaring both sides, we get $16 - p = 9 \Rightarrow p = 7$.

Answer: $7$