Answer
Center: $(-1,1)$
Foci: $(-1,0),(-1,2)$
Vertices: $(-1,-1),(-1,3)$
See graph
Work Step by Step
We are given the ellipse:
$4x^2+3y^2+8x-6y=5$
Put the equation in standard form:
$4(x^2+2x+1)-4+3(y^2-2y+1)-3=5$
$4(x+1)^2+3(y-1)^2=12$
$\dfrac{4(x+1)^2}{12}+\dfrac{3(y-1)^2}{12}=1$
$\dfrac{(x+1)^2}{3}+\dfrac{(y-1)^2}{4}=1$
The equation is in the form:
$\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$
Identify $h,k,a,b$:
$h=-1$
$k=1$
$a^2=4\Rightarrow a=\sqrt{4}=2$
$b^2=3\Rightarrow b=\sqrt 3$
Determine $c$:
$a^2=b^2+c^2$
$c^2=a^2-b^2$
$c^2=4-3$
$c^2=1$
$c=1$
Determine the center:
$(h,k)=(-1,1)$
Determine the foci:
$(h,k-c)=(-1,1-1)=(-1,0)$
$(h,k+c)=(-1,1+1)=(-1,2)$
Determine the vertices:
$(h,k-a)=(-1,1-2)=(-1,-1)$
$(h,k+a)=(-1,1+2)=(-1,3)$
Determine the co-vertices:
$(h-b,k)=(-1-\sqrt 3,1)$
$(h+b,k)=(-1+\sqrt 3,1)$
Graph the ellipse:
