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

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Answer 1 · 2017-01-27T11:54:29

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

The equation in the standard form is

$x^2/(sqrt2)^2-y^2/1^2=1$ , giving

center C(0, 0),

major axis A'A : x-axis,

asymptotes : $(x^2/2-y^2)=(x/sqrt2-y)(x/sqrt2+y)=0$ ,

$a = sqrt2, b = 1$ .

CS= CS'=ae=sqrt3,

$e = sqrt(1+b^2/a^2)=sqrt(3/2)$ ,

vertices $A(sqrt2, 0) and A'(-sqrt2, 0)$ and

foci : $S(sqrt3, 0) and S'(-sqrt3, 0).$

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]}

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