College Algebra, Chapter 8, 8.4, Section 8.4, Problem 6

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

The given ellipse is shifted so that its center is at $(3, -3)$. It is derived from the ellipse $\displaystyle \frac{x^2}{16} + y^2 = 1$ with center at the origin, by shifting it 3 units to the right and 3 units downward. The endpoints of the major and minor axis of the unshifted ellipse are $(4, 0), (-4, 0), (0, 1)$ and $(0, -1)$. We apply the required shifts to these points to obtain the corresponding points on the shifted ellipse:


$\begin{equation} \begin{aligned} (4, 0) \to (4 + 3, 0 -3) =& (7,-3) \\ \\ (-4, 0) \to (-4 +3, 0-3) =& (-1, -3) \\ \\ (0, 1) \to (0 + 3, 1 - 3) =& (3, -2) \\ \\ (0, -1) \to (0 + 3, -1-3) =& (3, -4) \end{aligned} \end{equation}$



To find the foci of the shifted ellipse, we first find the foci of the unshifted ellipse. Since $a^2 = 16$ and $b^2 = 1$, we have $c^2 = 16 - 1 = 15$, so $c = \sqrt{15}$. So the foci are $(\pm \sqrt{15}, 0)$. By applying transformations, we get


$\begin{equation} \begin{aligned} (\sqrt{15}, 0) \to (\sqrt{15} + 3, 0-3) =& (\sqrt{15} + 3, -3) \\ \\ (-\sqrt{15}, 0) \to (-\sqrt{15} + 3, 0-3) =& (-\sqrt{15} + 3, -3) \end{aligned} \end{equation}$


Thus, the foci of the shifted ellipse are


$(\sqrt{15} + 3, -3)$ and $(-\sqrt{15} + 3, -3)$

Therefore, its graph is