Answer
Vertices: $(0,-4),(0,4)$
Foci: $(0,-2\sqrt{3}),(0,2\sqrt{3})$
See graph
Work Step by Step
We are given the ellipse:
$4x^2+y^2=16$
Put the equation in standard form:
$\dfrac{4x^2}{16}+\dfrac{y^2}{16}=1$
$\dfrac{x^2}{4}+\dfrac{y^2}{16}=1$
The equation is in the form:
$\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$
Determine $h,k,a,b,c$:
$h=0$
$k=0$
$a^2=16\Rightarrow a=\sqrt{16}=4$
$b^2=4\Rightarrow b=\sqrt 4=2$
$c^2=a^2-b^2=16-4=12$
$c=2\sqrt{3}$
The center is:
$(h,k)=(0,0)$
Find the vertices:
$(h,k-a)=(0,0-4)=(0,-4)$
$(h,k+a)=(0,0+4)=(0,4)$
Find the foci:
$(h,k-c)=(0,0-2\sqrt{3})=(0,-2\sqrt{3})$
$(h,k+c)=(0,0+2\sqrt{3})=(0,2\sqrt{3})$
Use the value of $b=2$ to find the two points to the left and right of the center:
$(h-b,k)=(0-2,0)=(-2,0)$
$(h+b,k)=(0+2,0)=(2,0)$
Plot the center, the vertices, the two points above and below the center, and the foci. Then, graph the ellipse:
