The problem requires finding the value of \(p\) given that the foci of an ellipse and a hyperbola coincide. We need to analyze both equations, find their foci, and equate them.

Ellipse Foci Calculation

The equation of the ellipse is given as \(px^2 + 16y^2 = 16p\). To find the foci, we first standardize the equation by dividing by \(16p\): \( \frac{px^2}{16p} + \frac{16y^2}{16p} = \frac{16p}{16p} \) \( \frac{x^2}{16} + \frac{y^2}{p} = 1 \) This equation is in the standard form \(\frac{x^2}{a_e^2} + \frac{y^2}{b_e^2} = 1\). We have \(a_e^2 = 16\) and \(b_e^2 = p\). Given the constraint \(p < 16\), we know that \(a_e^2 > b_e^2\). This means the major axis of the ellipse lies along the x-axis. The distance from the center to the focus for an ellipse is calculated using \(c_e^2 = a_e^2 - b_e^2\). \( c_e^2 = 16 - p \) Thus, the foci of the ellipse are located at \((\pm \sqrt{16-p}, 0)\).

Hyperbola Foci Calculation

The equation of the hyperbola is given as \(25(81x^2 - 144y^2) = 11664\). First, simplify the equation by dividing both sides by 11664: \( \frac{25(81x^2)}{11664} - \frac{25(144y^2)}{11664} = 1 \) \( \frac{2025x^2}{11664} - \frac{3600y^2}{11664} = 1 \) Now, convert it into the standard form \(\frac{x^2}{a_h^2} - \frac{y^2}{b_h^2} = 1\). \( \frac{x^2}{11664/2025} - \frac{y^2}{11664/3600} = 1 \) \( \frac{x^2}{144/25} - \frac{y^2}{81/25} = 1 \) Here, \(a_h^2 = \frac{144}{25}\) and \(b_h^2 = \frac{81}{25}\). The transverse axis is along the x-axis. The distance from the center to the focus for a hyperbola is calculated using \(c_h^2 = a_h^2 + b_h^2\). \( c_h^2 = \frac{144}{25} + \frac{81}{25} = \frac{144 + 81}{25} = \frac{225}{25} = 9 \) \( c_h = \sqrt{9} = 3 \) Thus, the foci of the hyperbola are located at \((\pm 3, 0)\).

Finding Parameter p

The problem states that the foci of the ellipse and the hyperbola coincide. Therefore, their distances from the center must be equal.

Equating the foci distances: