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

1 Answer
Jan 27, 2017

Answer:

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/(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]}