Chapter 9 - Analytic Geometry - Cumulative Review - Review Exercises - Page 719: 16

$\frac{(x+4)^2}{16}+\frac{(y-5)^2}{25}=1$ See graph.

1. Based on the given conditions, the ellipse has a vertical major axis with center $(-4,5)$ (midpoint between the two foci), thus we have a general form $\frac{(x+4)^2}{b^2}+\frac{(y-5)^2}{a^2}=1$ 2. With $c=3, a=5$, we have $b^2=a^2-c^2=16$, thus the equation $\frac{(x+4)^2}{16}+\frac{(y-5)^2}{25}=1$ 3. See graph.