Answer
Center: $(0,-2)$
Foci: $(0,-2-\sqrt 3),(0,-2+\sqrt 3)$
Vertices: $(0,-4),(0,0)$
See graph
Work Step by Step
We are given the ellipse:
$4x^2+y^2+4y=0$
Put the equation in standard form:
$4x^2+(y^2+4y+4)-4=0$
$4x^2+(y+2)^2=4$
$\dfrac{4x^2}{4}+\dfrac{(y+2)^2}{4}=1$
$\dfrac{x^2}{1}+\dfrac{(y+2)^2}{4}=1$
The equation is in the form:
$\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$
Identify $h,k,a,b$:
$h=0$
$k=-2$
$a^2=4\Rightarrow a=\sqrt{4}=2$
$b^2=1\Rightarrow b=1$
Determine $c$:
$a^2=b^2+c^2$
$c^2=a^2-b^2$
$c^2=4-1$
$c^2=3$
$c=\sqrt 3$
Determine the center:
$(h,k)=(0,-2)$
Determine the foci:
$(h,k-c)=(0,-2-\sqrt 3)$
$(h,k+c)=(0,-2+\sqrt 3)$
Determine the vertices:
$(h,k-a)=(0,-2-2)=(0,-4)$
$(h,k+a)=(0,-2+2)=(0,0)$
Determine the co-vertices:
$(h-b,k)=(0-1,-2)=(-1,-2)$
$(h+b,k)=(0+1,-2)=(1,-2)$
Graph the ellipse:
