The foci of the ellipse px² + 16y² = 16p and the foci of the hyperbola 25(81x² − 144y²) = 11664 coincide (assume p < 16). What is the value of p?

Hyperbola: 25(81x² − 144y²) = 11664 → x²/(11664/(25·81)) − y²/(11664/(25·144)) = 1 → x²/(11664/2025) − y²/(11664/3600) = 1 = x²/(144/25) − y²/(81/25) = 1. So a² = 144/25, b² = 81/25, c² = a²(1+b²/a²) wait for hyperbola c² = a² + b² = 144/25 + 81/25 = 225/25 = 9, so c = 3. Foci at (±3, 0). Ellipse: px² + 16y² = 16p → x²/16 + y²/p = 1. Since p < 16, a² = 16, b² = p, c² = 16 − p. For foci to coincide: 16 − p = 9, so p = 7.