Answer
The vertices are $\left( { \pm a,0} \right) = \left( { \pm 2,0} \right)$. The foci are ${F_1} = \left( {\sqrt 6 ,0} \right)$ and ${F_2} = \left( { - \sqrt 6 ,0} \right)$.
Work Step by Step
We have ${x^2} - 2{y^2} = 4$
Divide both sides by 4 gives
$\frac{{{x^2}}}{4} - \frac{{{y^2}}}{2} = 1$, ${\ \ \ }$ ${\left( {\frac{x}{2}} \right)^2} - {\left( {\frac{y}{{\sqrt 2 }}} \right)^2} = 1$
By Theorem 2 of Section 12.5, this is the equation of a hyperbola in standard position, where $a=2$ and $b = \sqrt 2 $. The vertices are $\left( { \pm a,0} \right) = \left( { \pm 2,0} \right)$.
We have
$c = \sqrt {{a^2} + {b^2}} = \sqrt {4 + 2} = \sqrt 6 $
The foci are ${F_1} = \left( {\sqrt 6 ,0} \right)$ and ${F_2} = \left( { - \sqrt 6 ,0} \right)$.
