College Algebra, Chapter 8, Review Exercises, Section Review Exercises, Problem 32

Identify the type of curve which is represented by the equation $\displaystyle \frac{x^2}{12} + \frac{y^2}{144} = \frac{y}{12}$.
Find the foci and vertices(if any), and sketch the graph

$\begin{equation} \begin{aligned} 12x^2 + y^2 &= 12 y && \text{Multiply by } 144\\ \\ 12x^2 + y^2 - 12y &= 0 && \text{Subtract }12y \\ \\ 12x^2 + y^2 - 12y + 36 &= 36 && \text{Complete the square: Add } \left( \frac{-12}{2} \right)^2 = 36\\ \\ 12x^2 + (y - 6)^2 &= 36 && \text{Perfect square}\\ \\ \frac{x^2}{3} + \frac{(y-6)^2}{36} &= 1 && \text{Divide by } 36 \end{aligned} \end{equation}$

Since the equation is a sum of the squares, then it is an ellipse. The shifted ellipse has the form $\displaystyle \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1$
with center at $(h,k)$ and vertical major axis. It is derived from the ellipse $\displaystyle \frac{x^2}{3} + \frac{y^2}{36} = 1$ with center at origin by
shifting it $6$ units upward. Thus, by applying transformation

$\begin{equation} \begin{aligned} \text{center } & (h,k) && \rightarrow && (0,6)\\ \\ \text{vertices:major axis } & (0,a)&& \rightarrow && (0,6) && \rightarrow && (0,6+6) && = && (0,12)\\ \\ & (0,-a)&& \rightarrow && (0,-6) && \rightarrow && (0,-6+6) && = && (0,0)\\ \\ \text{minor axis } & (b,0)&& \rightarrow && (\sqrt{3},0) && \rightarrow && (\sqrt{3},0+6) && = && (\sqrt{3},6)\\ \\ & (-b,0)&& \rightarrow && (-\sqrt{3},0) && \rightarrow && (-\sqrt{3},0+6) && = && (\sqrt{3},6) \end{aligned} \end{equation}$

The foci of the unshifted ellipse are determined by $c = \sqrt{a^2 - b^2} = \sqrt{3c - 3} = \sqrt{33}$
Thus, by applying transformation

$\begin{equation} \begin{aligned} \text{minor axis } & (0,c)&& \rightarrow && (0,\sqrt{33}) && \rightarrow && (0,\sqrt{33}+6) && = && (0,6+\sqrt{33})\\ \\ & (0,-c)&& \rightarrow && (0,-\sqrt{33}) && \rightarrow && (0,\sqrt{33}+6) && = && (0,6-\sqrt{33}) \end{aligned} \end{equation}$

Therefore, the graph is