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

Find the center, foci and vertices of the ellipse $\displaystyle \frac{(x + 2)^2}{4} + y^2 = 1$ and determine the lengths of the major and minor axes. Sketch its graph.

The given ellipse is shifted so that its center is at $(-2, 0)$. It is derived from the ellipse $\displaystyle \frac{x^2}{4} + y^2 = 1$ with center at the origin, by shifting it 2 units to the left. The endpoints of the major and minor axis of the unshifted ellipse are $(2, 0), (-2, 0), (0, 1)$ and $(0, -1)$. By applying transformations, we get the corresponding points of the shifted ellipse:


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$



To find the foci of the shifted ellipse, we first find the foci of the unshifted ellipse. Since $a^2 = 4$ and $b^2 = 1$, we have $c^2 = 4 - 1 = 3$, so $c = \sqrt{3}$. So the foci are $(\pm \sqrt{3}, 0)$. Again, by applying transformations, we get


$
\begin{equation}
\begin{aligned}

(\sqrt{3}, 0) \to (\sqrt{3} - 2, 0) =& (\sqrt{3} - 2, 0)
\\
\\
(-\sqrt{3}, 0) \to (-\sqrt{3} - 2, 0) =& (-\sqrt{3} - 2, 0)

\end{aligned}
\end{equation}
$


Thus, the foci of the shifted ellipse are


$(\sqrt{3} - 2, 0)$ and $(-\sqrt{3} - 2, 0)$

Therefore, its graph is