Answer
Center: $(1,-2)$
Foci: $(1,-2-\sqrt 5),(1,-2+\sqrt 5)$
Vertices: $(1,-5),(1,1)$
See graph
Work Step by Step
We are given the ellipse:
$9x^2+4y^2-18x+16y-11=0$
Put the equation in standard form:
$9(x^2-2x+1)-9+4(y^2+4y+4)-16-11=0$
$9(x-1)^2+4(y+2)^2=36$
$\dfrac{9(x-1)^2}{36}+\dfrac{4(y+2)^2}{36}=1$
$\dfrac{(x-1)^2}{4}+\dfrac{(y+2)^2}{9}=1$
The equation is in the form:
$\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$
Identify $h,k,a,b$:
$h=1$
$k=-2$
$a^2=9\Rightarrow a=\sqrt{9}=3$
$b^2=4\Rightarrow b=\sqrt 4=2$
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:
$(h,k)=(1,-2)$
Determine the foci:
$(h,k-c)=(1,-2-\sqrt 5)$
$(h,k+c)=(1,-2+\sqrt 5)$
Determine the vertices:
$(h,k-a)=(1,-2-3)=(1,-5)$
$(h,k+a)=(1,-2+3)=(1,1)$
Determine the co-vertices:
$(h-b,k)=(1-2,-2)=(-1,-2)$
$(h+b,k)=(1+2,-2)=(3,-2)$
Graph the ellipse:
