Chapter 9 - Analytic Geometry - Section 9.3 The Ellipse - 9.3 Assess Your Understanding - Page 677: 32

$\frac{x^2}{12}+\frac{y^2}{16}=1$ See graph.

1. Based on the given conditions, we have $c=2, a=8/2=4$, center $(0,0)$ with a vertical major axis, thus $b=\sqrt {a^2-c^2}=\sqrt {16-4}=2\sqrt 3$, and $\frac{x^2}{12}+\frac{y^2}{16}=1$ 2. See graph.