Answer
The foci are $(3,\pm 4\sqrt{2}-1)$ and the vertices are $(3,-7)$ and $(3,5)$.
Work Step by Step
Recall: The ellipse $\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$ for $a\geq b$ has the foci at $(h,\pm c+k)$ and the vertices at $(h,\pm a+k)$.
We have
$9x^2-54x+y^2+2y+46=0$
$9x^2-54x+81+y^2+2y+1+46=81+1$
$9(x^2-6x+9)+(y+1)^2+46=82$
$9(x-3)^2+(y+1)^2=36$
$\frac{9(x-3)^2}{36}+\frac{(y+1)^2}{36}=1$
$\frac{(x-3)^2}{2^2}+\frac{(y-(-1))^2}{6^2}=1$
Then,
$h=3$, $k=-1$, $b=2$, and $a=6$
$c=\sqrt{a^2-b^2}=\sqrt{6^2-2^2}=\sqrt{36-4}=\sqrt{32}=4\sqrt{2}$
So, the foci are $(3,\pm 4\sqrt{2}-1)$ and the vertices are $(3,-7)$ and $(3,5)$.
