Answer
Center: $(1,0)$
Foci: $(1,-2\sqrt 2),(1,2\sqrt 2)$
Vertices: $(1,-3),(1,3)$
See graph
Work Step by Step
We are given the ellipse:
$9x^2+y^2-18x=0$
Put the equation in standard form:
$9(x^2-2x+1)-9+y^2=0$
$9(x-1)^2+y^2=9$
$\dfrac{9(x-1)^2}{9}+\dfrac{y^2}{9}=1$
$\dfrac{(x-1)^2}{1}+\dfrac{y^2}{9}=1$
The equation is in the form:
$\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$
Identify $h,k,a,b$:
$h=1$
$k=0$
$a^2=9\Rightarrow a=\sqrt{9}=3$
$b^2=1\Rightarrow b=1$
Determine $c$:
$a^2=b^2+c^2$
$c^2=a^2-b^2$
$c^2=9-1$
$c^2=8$
$c=\sqrt 8=2\sqrt 2$
Determine the center:
$(h,k)=(1,0)$
Determine the foci:
$(h,k-c)=(1,0-2\sqrt 2)=(1,-2\sqrt 2)$
$(h,k+c)=(1,0+2\sqrt 2)=(1,2\sqrt 2)$
Determine the vertices:
$(h,k-a)=(1,0-3)=(1,-3)$
$(h,k+a)=(1,0+3)=(1,3)$
Determine the co-vertices:
$(h-b,k)=(1-1,0)=(0,0)$
$(h+b,k)=(1+1,0)=(2,0)$
Graph the ellipse:
