College Algebra, Chapter 8, 8.4, Section 8.4, Problem 10

Find the center, foci, vertices and asymptotes of the hyperbola $\displaystyle (y + 5)^2 = -6x + 12$. Sketch its graph.

We can rewrite the equation as $(y + 5)^2 = -6 (x - 2)$. This parabola opens to the left with vertex at $(2, -5)$. It is obtain from the parabola $y^2 = -6x$ by shifting 2 units to the right and 5 units downward. Since $4p = 6$, we have $\displaystyle p = \frac{3}{2}$. So the focus is $\displaystyle \frac{3}{2}$ units from the left of the vertex and the directrix is $\displaystyle \frac{3}{2}$ units to the right of the vertex.

Therefore, the focus is at

$\displaystyle (2, -5) \to \left( 2 - \frac{3}{2}, -5 \right) = \left( \frac{1}{2}, -5 \right)$

and the directrix is the line

$\displaystyle x = 2 + \frac{3}{2} = \frac{7}{2}$