College Algebra, Chapter 8, 8.4, Section 8.4, Problem 26

Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, find the center, foci, vertices and lengths of the major and minor axes. If it is a parabola, find the vertex, focus and directrix. If it is a hyperbola, find the center, foci, vertices and asymptotes. Sketch the graph of the equation. If the equation has no graph, explain why.


$\begin{equation} \begin{aligned} x^2 + 6x + 12y + 9 =& 0 && \text{Subtract $9$ and $12y$} \\ \\ x^2 + 6x =& -12y - 9 && \text{Complete the square: add } \left( \frac{6}{2} \right)^2 = 9 \\ \\ x^2 + 6x + 9 =& -12y-9+9 && \text{Perfect square} \\ \\ (x + 3)^2 =& -12y && \end{aligned} \end{equation}$


The equation has the form $(x - h)^2 = -4py$, it is a parabola that has vertex on $(-3, 0)$ and opens downward. It is obtain from the parabola $x^2 = -12y$ by shifting it $3$ units to the left. Since $4p = 12$, we have $p = 3$. So the focus is $3$ units below and the vertex and the directrix is $3$ units above the vertex.

Therefore, the focus is at

$(-3, 0) \to (-3, 0 - 3 ) = (-3, -3)$

and the directrix is the line $y = 0 + 3 = 3$