Chapter 9 - Analytic Geometry - Section 9.3 The Ellipse - 9.3 Assess Your Understanding - Page 677: 36

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

1. Based on the given conditions, we have $a=4, b=1$, center $(0,0)$ with a horizontal major axis, thus $c=\sqrt {a^2-b^2}=\sqrt {16-1}=\sqrt {15}$, and $\frac{x^2}{16}+\frac{y^2}{1}=1$ 2. See graph.