Answer
Center: $(0,2)$
Foci: $(-\sqrt 2,2),(\sqrt 2,2)$
Vertices: $(-\sqrt 3,2),(\sqrt 3,2)$
See graph
Work Step by Step
We are given the ellipse:
$x^2+3y^2-12y+9=0$
Put the equation in standard form:
$x^2+3(y^2-4y+4)-12+9=0$
$x^2+3(y-2)^2=3$
$\dfrac{x^2}{3}+\dfrac{3(y-2)^2}{3}=1$
$\dfrac{x^2}{3}+\dfrac{(y-2)^2}{1}=1$
The equation is in the form:
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
Identify $h,k,a,b$:
$h=0$
$k=2$
$a^2=3\Rightarrow a=\sqrt 3$
$b^2=1\Rightarrow b=1$
Determine $c$:
$a^2=b^2+c^2$
$c^2=a^2-b^2$
$c^2=3-1$
$c^2=2$
$c=\sqrt 2$
Determine the center:
$(h,k)=(0,2)$
Determine the foci:
$(h-c,k)=(0-\sqrt 2,2)=(-\sqrt 2,2)$
$(h+c,k)=(0+\sqrt 2,2)=(\sqrt 2,2)$
Determine the vertices:
$(h-a,k)=(0-\sqrt 3,2)=(-\sqrt 3,2)$
$(h+a,k)=(0+\sqrt 3,2)=(\sqrt 3,2)$
Determine the $y$-intercepts:
$(h,k-b)=(0,2-1)=(0,1)$
$(h,k+b)=(0,2+1)=(0,3)$
Graph the ellipse:
