Answer
Center: $(-2,1)$
Foci: $(-2-\sqrt 3,1),(-2+\sqrt 3,1)$
Vertices: $(-4,1),(0,1)$
See graph
Work Step by Step
We are given the ellipse:
$x^2+4x+4y^2-8y+4=0$
Put the equation in standard form:
$(x^2+4x+4)-4+4(y^2-2y+1)-4+4=0$
$(x+2)^2+4(y-1)^2=4$
$\dfrac{(x+2)^2}{4}+\dfrac{4(y-1)^2}{4}=1$
$\dfrac{(x+2)^2}{4}+\dfrac{(y-1)^2}{1}=1$
The equation is in the form:
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
Identify $h,k,a,b$:
$h=-2$
$k=1$
$a^2=4\Rightarrow a=\sqrt 4=2$
$b^2=1\Rightarrow b=1$
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:
$(h,k)=(-2,1)$
Determine the foci:
$(h-c,k)=(-2-\sqrt 3,1)$
$(h+c,k)=(-2+\sqrt 3,1)$
Determine the vertices:
$(h-a,k)=(-2-2,1)=(-4,1)$
$(h+a,k)=(-2+2,1)=(0,1)$
Determine the $y$-intercepts:
$(h,k-b)=(-2,1-1)=(-2,0)$
$(h,k+b)=(-2,1+1)=(-2,2)$
Graph the ellipse:
