## Calculus (3rd Edition)

- the vertices are $(\pm 4,0)$ and $(0, \pm 2)$ - the foci are $(0, \pm \sqrt{12})$ - the center is $(0,0)$
The equation $4x^2+ y^2=16$ can be written in the form $$\left(\frac{x}{2}\right)^{2}+\left(\frac{y}{4}\right)^{2}=1$$ which is an ellipse with $a=2, b=4$ so $b\gt a$ and hence $c=\sqrt{b^2-a^2}=\sqrt{16-4}=\sqrt{12}$. So, we have: the vertices are $(0,\pm 4)$ and $(\pm 2,0)$ the foci are $(0, \pm \sqrt{12})$ the center is $(0,0)$