## Calculus: Early Transcendentals 8th Edition

Hyperbola Foci: $( \pm \sqrt 5,0)$ Vertices: $(\pm1,0,)$
$4x^2=y^2+4$ $\frac{x^{2}}{1^2}-\frac{y^{2}}{2^2}=1$, the equation of a hyperbola. $c^{2}=1+4=5$ $c=\sqrt 5$ Foci: $( \pm \sqrt 5,0)$ Vertices: $(\pm1,0,)$