Answer
The vertices are $\left( { \pm \frac{1}{{\sqrt 2 }},3} \right)$.
The foci are ${F_1} = \left( {\sqrt {\frac{3}{2}} ,3} \right)$ and ${F_2} = \left( { - \sqrt {\frac{3}{2}} ,3} \right)$.
Work Step by Step
Write
${\left( {y - 3} \right)^2} = 2{x^2} - 1$
$2{x^2} - {\left( {y - 3} \right)^2} = 1$
${\left( {\frac{x}{{1/\sqrt 2 }}} \right)^2} - {\left( {\frac{{y - 3}}{1}} \right)^2} = 1$
By Theorem 2 of Section 12.5, this is the equation of a hyperbola centered at $\left( {0,3} \right)$, where $a = \frac{1}{{\sqrt 2 }}$ and $b=1$. So, the vertices are $\left( { \pm a,0} \right) + \left( {0,3} \right) = \left( { \pm \frac{1}{{\sqrt 2 }},3} \right)$.
We have
$c = \sqrt {{a^2} + {b^2}} = \sqrt {\frac{1}{2} + 1} = \sqrt {\frac{3}{2}} $
The foci are $\left( { \pm c,0} \right) + \left( {0,3} \right) = \left( { \pm \sqrt {\frac{3}{2}} ,3} \right)$. So, ${F_1} = \left( {\sqrt {\frac{3}{2}} ,3} \right)$ and ${F_2} = \left( { - \sqrt {\frac{3}{2}} ,3} \right)$.
