Chapter 10 - Parametric Equations and Polar Coordinates - 10.5 Conic Sections - 10.5 Exercises - Page 720: 25

Foci: $( \pm \sqrt 5,0)$ and vertices: $(\pm1,0,)$

Given: $4x^2=y^2+4$ It can be rewritten as: $\frac{x^{2}}{1^2}-\frac{y^{2}}{2^2}=1$ This represents the equation of an hyperbola with $c^{2}=1+4=5$ or, $c=\sqrt 5$ Hence, Foci: $( \pm \sqrt 5,0)$ and vertices: $(\pm1,0)$