Answer
Ellipse
Vertices are (1 +/- $\sqrt 2$, 2)
Foci are (0,2), (2,2)
Work Step by Step
$x^{2}$ - 2x + $2y^{2}$ - 8y + 7 = 0
$x^{2}$ - 2x + $2y^{2}$ - 8y = -7
$x^{2}$ - 2x +1 + $2(y^{2}-4y+4)$ = -7 + 1 + 2(4)
$(x-1)^{2}$ + 2$(y-2)^{2}$ = -7 + 1 + 8 = 2
$(x-1)^{2}$ + 2$(y-2)^{2}$ = 2
$\frac{(x-1)^{2}}{2}$ + $\frac{(y-2)^{2}}{1}$ = 1
This is an ellipse; the center is at (1,2)
The vertices are (1 +/- $\sqrt 2$, 2)
For the foci, we know from the equation of the ellipse that $a^{2}$ = 2; $b^{2}$ = 1
$c^{2}$ = $a^{2}$ - $b^{2}$
$c^{2}$ = 2-1
$c^{2}$ = 1
c = 1
The Foci are (0,2), (2,2)
