.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^2 + 6y + .... + 12x = 39
y^2 + 6y + 9 + 12x = 39+9
(y+3)^2 - 48 = -12x
x = -(1/12)(y+3)^2 + 4

so your parabola has centre at (4,-3), opening to the left
and from -4px = -12x, we see that p = 3
and the focus would be (1,-3) and directrix x = 7

so the centre of the hyperbola is (7,-3)
from my sketch, a = 3, c = 6
then a^2 + b^2 = c^2
9 + b^2 = 36
b^2 = 27

my equation would be
(x-7)^2 / 9 + (y + 3)^2 / 27 = 1

graphing this on
www.desmos.com/calculator
shows that it is correct