## Precalculus (10th Edition)

Center: $(0,2)$ Foci: $(-\sqrt 2,2),(\sqrt 2,2)$ Vertices: $(-\sqrt 3,2),(\sqrt 3,2)$ See graph
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: