# How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola x^2-2y^2=2?

Jan 27, 2017

See data in the explanation and the illustrative graph that marks the foci and the asymptotes..

#### Explanation:

The equation in the standard form is

${x}^{2} / {\left(\sqrt{2}\right)}^{2} - {y}^{2} / {1}^{2} = 1$, giving

center C(0, 0),

major axis A'A : x-axis,

asymptotes : $\left({x}^{2} / 2 - {y}^{2}\right) = \left(\frac{x}{\sqrt{2}} - y\right) \left(\frac{x}{\sqrt{2}} + y\right) = 0$,

$a = \sqrt{2} , b = 1$.

CS= CS'=ae=sqrt3,

$e = \sqrt{1 + {b}^{2} / {a}^{2}} = \sqrt{\frac{3}{2}}$,

vertices $A \left(\sqrt{2} , 0\right) \mathmr{and} A ' \left(- \sqrt{2} , 0\right)$ and

foci : $S \left(\sqrt{3} , 0\right) \mathmr{and} S ' \left(- \sqrt{3} , 0\right) .$

graph{(x^2/2-y^2-1)((x-1.73)^2+y^2-.01)((x+1.73)^2+y^2-.01)(x^2-y^2)=0^2 [-10, 10, -5, 5]}