Chapter 9 - Analytic Geometry - Section 9.4 The Hyperbola - 9.4 Assess Your Understanding - Page 689: 27

$\frac{x^2}{8}-\frac{y^2}{8}=1$ See graph.

1. Given foci $(-4,0)$ and $(4,0)$, asymptote $y=-x$, we have a horizontal transverse axis, center $(0,0)$ (midpoint of foci), $c=4, \frac{b}{a}=1$, $a^2+b^2=c^2=16$, 2. Thus $a^2=b^2=8$ and the equation is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ or $\frac{x^2}{8}-\frac{y^2}{8}=1$ 3. See graph.