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