Answer
$$\eqalign{
& {\bf{e}} = \sqrt {10} \cr
& {\bf{e}} \approx 3.16 \cr} $$
Work Step by Step
$$\eqalign{
& \frac{{{x^2}}}{2} - \frac{{{y^2}}}{{18}} = 1 \cr
& {\text{The equation is written in the form }}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& \frac{{{x^2}}}{2} - \frac{{{y^2}}}{{18}} = 1,{\text{ then }}{a^2} = 2{\text{ and }}{b^2} = 18 \cr
& {\text{The excentricity of a hyperbola is giving by}} \cr
& {\bf{e}} = \frac{{\sqrt {{a^2} + {b^2}} }}{a} \cr
& {\bf{e}} = \frac{{\sqrt {2 + 18} }}{{\sqrt 2 }} \cr
& {\bf{e}} = \sqrt {\frac{{20}}{2}} \cr
& {\bf{e}} = \sqrt {10} \cr
& {\bf{e}} \approx 3.16 \cr} $$
