College Algebra, Chapter 8, Review Exercises, Section Review Exercises, Problem 14

Determine the center, vertices, foci, eccentricity and lengths of the major and minor axes of the ellipse $\displaystyle \frac{(x - 2)^2}{25} + \frac{(y + 3)^2}{16} = 1$. Then sketch its graph.

The shifted ellipse has the form $\displaystyle \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$ with center at $(h,k)$. It is derived from the ellipse $\displaystyle \frac{x^2}{25} + \frac{y^2}{16} = 1$ with center at origin, by shifting it $2$ units to the right and $3$ units downward. The endpoints of the vertices of the unshifted ellipse are $(\pm 5, 0)$ and $(0, \pm 4)$. By applying transformations, the corresponding vertices of the shifted ellipse are..


$
\begin{equation}
\begin{aligned}

(5, 0) \to (5 + 2, 0 - 3) =& (7, -3)
\\
\\
(-5, 0) \to (-5 + 2, 0 - 3) =& (-3, -3)
\\
\\
(0, 4) \to (0 + 2, 4 - 3) =& (2, 1)
\\
\\
(0, -4) \to (0 + 2, -4 - 3) =& (2, -7)

\end{aligned}
\end{equation}
$



Now, the foci of the unshifted ellipse is determined by $c = \sqrt{a^2 - b^2} = \sqrt{25 - 16} = 3$. Then by applying transformations, the foci of the shifted ellipse are..


$
\begin{equation}
\begin{aligned}

(3, 0) \to (3 + 2, 0 - 3) =& (5, -3)
\\
\\
(-3, 0) \to (-3 + 2, 0 - 3) =& (-1, -3)

\end{aligned}
\end{equation}
$


To sum it up,

eccentricity $\displaystyle \frac{c}{a} \to \frac{3}{5}$

length of the major axis $2a \to 10$

length of the minor axis $\displaystyle 2b \to 8$

Therefore, the graph is