Answer
- the vertices are $(\pm 4,0)$ and $(0, \pm 2)$
- the foci are $(0, \pm \sqrt{12})$
- the center is $(0,0)$
Work Step by Step
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 $(0,0)$
