Answer
Center: $(2,-1)$
Foci: $(1,-1),(3,-1)$
Vertices: $(2-\sqrt 3,-1),(2+\sqrt 3,-1)$
See graph
Work Step by Step
We are given the ellipse:
$2x^2+3y^2-8x+6y+5=0$
Put the equation in standard form:
$2(x^2-4x+4)-8+3(y^2+2y+1)-3+5=0$
$2(x-2)^2+3(y+1)^2=6$
$\dfrac{2(x-2)^2}{6}+\dfrac{3(y+1)^2}{6}=1$
$\dfrac{(x-2)^2}{3}+\dfrac{(y+1)^2}{2}=1$
The equation is in the form:
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
Identify $h,k,a,b$:
$h=2$
$k=-1$
$a^2=3\Rightarrow a=\sqrt{3}$
$b^2=2\Rightarrow b=\sqrt {2}$
Determine $c$:
$a^2=b^2+c^2$
$c^2=a^2-b^2$
$c^2=3-2$
$c^2=1$
$c=1$
Determine the center:
$(h,k)=(2,-1)$
Determine the foci:
$(h-c,k)=(2-1,-1)=(1,-1)$
$(h+c,k)=(2+1,-1)=(3,-1)$
Determine the vertices:
$(h-a,k)=(2-\sqrt 3,-1)$
$(h+a,k)=(2+\sqrt 3,-1)$
Determine the $y$-intercepts:
$(h,k-b)=(2,-1-\sqrt 2)$
$(h,k+b)=(2,-1+\sqrt 2)$
Graph the ellipse:
