College Algebra, Chapter 8, Review Exercises, Section Review Exercises, Problem 16

Determine the center, vertices, foci, eccentricity and lengths of the major and minor axes of the ellipse $\displaystyle 2x^2 + y^2 = 2 + 4(x - y)$. Then sketch its graph.


$
\begin{equation}
\begin{aligned}

2x^2 + y^2 =& 2 + 4x - 4y
&& \text{Distribute } 4
\\
\\
2(x^2 - 2x + \quad ) + (y^2 + 4y + \quad ) =& 2
&& \text{Group terms and factor}
\\
\\
2(x^2 - 2x + 1) + (y^2 + 4y + 4) =& 2 + 2 + 4
&& \text{Complete the square: add } \left( \frac{-2}{2} \right)^2 = 1 \text{ and } \left( \frac{4}{2} \right)^2 = 4 \text{ on the left side, then add 2 and 4 on the right side}
\\
\\
2(x - 1)^2 + (y + 2)^2 =& 8
&& \text{Perfect Square}
\\
\\
\frac{(x - 1)^2}{4} + \frac{(y + 2)^2}{8} =& 1
&& \text{Divide both sides by } 8

\end{aligned}
\end{equation}
$


The shifted ellipse now has the form $\displaystyle \frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1$ with center at $(h,k)$. It is derived from the ellipse $\displaystyle \frac{x^2}{4} + \frac{y^2}{8}$ with center at origin, by shifting it $1$ unit to the right and $2$ units downward. By applying transformations, the endpoints of the vertices of the shifted ellipse are..


$
\begin{equation}
\begin{aligned}

(0, a) \to (0, 2 \sqrt{2}) \to (0+1, 2 \sqrt{2} - 2) =& (1, 2 \sqrt{2} - 2)
\\
\\
(0, -a) \to (0, -2 \sqrt{2}) \to (0+1, -2 \sqrt{2} - 2) =& (1, - 2 \sqrt{2} - 2)
\\
\\
(b, 0) \to (2, 0) \to (2+1, 0 -2) =& (3,-2)
\\
\\
(-b, 0) \to (-2, 0) \to (-2+1, 0 -2) =& (-1,-2)

\end{aligned}
\end{equation}
$



Now, the foci of the unshifted ellipse are determined by $c = \sqrt{a^2 - b^2} = \sqrt{8-4} = 2$. Then by applying transformations, the foci of the shifted ellipse are..


$
\begin{equation}
\begin{aligned}

(0,c) \to (0,2) \to (0 + 1,2-2) =& (1,0)
\\
\\
(0,-c) \to (0,-2) \to (0 + 1,-2-2) =& (1,-4)

\end{aligned}
\end{equation}
$


To sum it up,

eccentricity $\displaystyle \frac{c}{a} \to \frac{\sqrt{2}}{2}$

length of the major axis $2a \to 4 \sqrt{2}$

length of the minor axis $\displaystyle 2b \to 4$

Therefore, the graph is