Answer
Center: $(3,-1)$
Foci: $(3,-1-\sqrt 5),(3,-1+\sqrt 5)$
Vertices: $(3,-4),(3,2)$
See graph
Work Step by Step
We are given the ellipse:
$\dfrac{(x-3)^2}{4}+\dfrac{(y+1)^2}{9}=1$
The equation is in the form:
$\dfrac{(x-h)^2}{b^2}+\dfrac{(y+1)^2}{a^2}=1$
Identify $h,k,a,b$:
$h=3$
$k=-1$
$a^2=9\Rightarrow a=\sqrt 9=3$
$b^2=4\Rightarrow b-=\sqrt 4=2$
Determine $c$:
$a^2=b^2+c^2$
$c^2=a^2-b^2$
$c^2=9-4$
$c^2=5$
$c=\sqrt 5$
Determine the center:
$(h,k)=(3,-1)$
Determine the foci:
$(h,k-c)=(3,-1-\sqrt 5)$
$(h,k+c)=(3,-1+\sqrt 5)$
Determine the vertices:
$(h,k-a)=(3,-1-3)=(3,-4)$
$(h,k+a)=(3,-1+3)=(3,2)$
Determine the $x$-intercepts:
$(h-b,k)=(3-2,-1)=(1,-1)$
$(h+b,k)=(3+2,-1)=(5,-1)$
Graph the ellipse:
