College Algebra, Chapter 8, Review Exercises, Section Review Exercises, Problem 42

Identify the type of curve which is represented by the equation $\displaystyle x^2 + 4y^2 = 4x + 8$
Find the foci and vertices(if any), and sketch the graph

$\begin{equation} \begin{aligned} x^2 - 4x + 4y^2 &= 8 && \text{Subtract } 4x\\ \\ x^2 - 4x + 4 + 4y^2 &= 8 + 4 && \text{Complete the square; Add } \left( \frac{-4}{2} \right)^2 = 4\\ \\ (x - 2)^2 + 4y^2 &= 12 && \text{Perfect square}\\ \\ \frac{(x-2)^2}{12} + \frac{y^2}{3} &= 1 && \text{Divide by 12} \end{aligned} \end{equation}$


The equation is an ellipse that has form $\displaystyle \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ with center at $(h,k)$ and horizontal major axis.
Since, the denominator of $x^2$ is bigger. The graph of the shifted ellipse is obtained by shifting the graph of $\displaystyle \frac{x^2}{12} + \frac{y^2}{3} =1$
by 2 units to the right. This gives us $a^2 = 12$ and $b^2 = 3$. So, $a = 2\sqrt{3}, b = \sqrt{3}$ and $c = \sqrt{a^2 - b^2} = \sqrt{12-3} = 3$. Thus,
by applying transformatioms, we have

$\begin{equation} \begin{aligned} \text{center } & (h,k) && \rightarrow && (2,0)\\ \\ \text{vertices: major axis} & (a,0)&& \rightarrow && (2\sqrt{3},0) && \rightarrow && (2\sqrt{3}+2,0) && = && (2\sqrt{3}+2,0)\\ \\ & (-a,0)&& \rightarrow && (-2\sqrt{3},0) && \rightarrow && (-2\sqrt{3}+2,0) && = && (-2\sqrt{3}+2,0)\\ \\ \text{minor axis }& (0,b)&& \rightarrow && (0,\sqrt{3}) && \rightarrow && (0+2, \sqrt{3}) && = && (2, \sqrt{3})\\ \\ & (0,-b)&& \rightarrow && (0,-\sqrt{3}) && \rightarrow && (0+2, -\sqrt{3}) && = && (2, - \sqrt{3})\\ \\ \text{foci }& (c,0)&& \rightarrow && (3,0) && \rightarrow && (3+2,0) && = && (5,0)\\ \\ & (-c,0)&& \rightarrow && (-3,0) && \rightarrow && (-3+2,0) && = && (-1,0) \end{aligned} \end{equation}$

Therefore, the graph is