Solving Problems Involving Different Conic Sections

Question

Find the standard equation of the hyperbola whose conjugate axis is on the directrix of the parabola 𝑦^2 + 12𝑥 + 6𝑦 = 39, having the focus of the parabola as one of its foci, and the vertex of the parabola as one of its vertices.

Answer 1

(y + 3)² = - 12x + 48; or (y + 3)² = - 12(x - 4). So, vertex of parabola (4, - 3); 4p = - 12; p = - 3.

Focus of parabola: (4 - 3, - 3) = (1, -3). Directrix of parabola x = 4 - (- 3) or x = 7.

Center of hyperbola at point (7, - 3). a = 3; c = 6; b² = 36 - 9 = 27.

Standard equation of hyperbola: (x - 7)²/9 - (y + 3)²/27 = 1