Answer
$${\text{The equation describes a hyperbola}}{\text{.}}$$
Work Step by Step
$$\eqalign{
& {x^2} - \frac{{{y^2}}}{2} = 1 \cr
& \left( a \right) \cr
& {\text{The equation is written in the form }}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& {\text{then, the equation describes a hyperbola}}{\text{.}} \cr
& \underbrace {{x^2} - \frac{{{y^2}}}{2} = 1}_{\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1} \to a = 1,{\text{ }}b = \sqrt 2 \cr
& c = \sqrt {{a^2} + {b^2}} = \sqrt {1 + 2} = \sqrt 3 \cr
& \cr
& \left( b \right) \cr
& {\text{Foci: }}\left( { - c,0} \right){\text{ and }}\left( {c,0} \right) \cr
& {\text{Foci: }}\left( { - \sqrt 3 ,0} \right){\text{ and }}\left( {\sqrt 3 ,0} \right) \cr
& {\text{Vertices: }}\left( { - a,0} \right){\text{ and }}\left( {a,0} \right) \cr
& {\text{Vertices: }}\left( { - 1,0} \right){\text{ and }}\left( {1,0} \right) \cr
& \cr
& \left( c \right) \cr
& {\text{Eccentricity }} \cr
& e = \frac{c}{a} \cr
& e = \sqrt 3 \cr
& \cr
& \left( d \right) \cr
& {\text{Graph}} \cr} $$
