College Algebra, Chapter 8, Review Exercises, Section Review Exercises, Problem 34

Identify the type of curve which is represented by the equation$\displaystyle 32 x^2 + 6 x = 9y^2$
Find the foci and vertices(if any), and sketch the graph

$\begin{equation} \begin{aligned} x^2 + 6x - 9y^2 &= 0 && \text{Subtract } 9y^2\\ \\ x^2 + 6x + 9 - 9y^2 &= 9 && \text{Complete the square: Add } \left( \frac{6}{2} \right)^2 = 9\\ \\ (x + 3)^2 - 9y^2 &= 9 && \text{Perfect Square}\\ \\ \frac{(x+3)^2}{9} - y^2 &= 1 && \text{Divide by 9} \end{aligned} \end{equation}$

The equation is hyperbola that has the form $\displaystyle \frac{(x-h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$ with center at $(h,k)$ and horizontal transverse axis
because the denominator of the $x^2$ is positive. Thus gives $a^2 = 9$ and $b^2 = 1$, so $a = 3, b =1$ and $c = \sqrt{a^2 + b^2} = \sqrt{9+1} = \sqrt{10}$. The
graph of the shifted ellipse is obtained by shifting the graph of $\displaystyle \frac{x^2}{9} - y^2 = 1$ three units to the left. Thus by applying transformation

$\begin{equation} \begin{aligned} \text{center } & (h,k) && \rightarrow && (-3,0)\\ \\ \text{vertices} & (a,0)&& \rightarrow && (3,0) && \rightarrow && (3-3,0) && = && (0,0)\\ \\ & (-a,0)&& \rightarrow && (-3,0) && \rightarrow && (-3-3,0) && = && (-6,0)\\ \\ \text{foci } & (c,0)&& \rightarrow && (\sqrt{10},0) && \rightarrow && (\sqrt{10}-3,0) && = && (-3+\sqrt{10},6)\\ \\ & (-c,0)&& \rightarrow && (-\sqrt{10},0) && \rightarrow && (-\sqrt{10}-3,0) && = && (-3-\sqrt{10},0) \end{aligned} \end{equation}$

Therefore, the graph is