How do you write the equation of the hyperbola given Foci: (-3,0),(3,0) and vertices (-1,0), (1,0)?

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Answer 1 · 2018-01-19T18:10:42

# x^2-y^2/8=1, or, 8x^2-y^2=8#.

The focii #(+-3,0)# lie on the #X"-Axis"#, hence, the Transverse

axis of the hyperbola S is #X"-Axis"#.

Its eqn. is given by, #S : x^2/a^2-y^2/b^2=1#.

For #S#, the focii are #(+-ae,0)#, and the vertices are #(+-a,0)#.

(e eccentricity)

#:. ae=3, and, a=1 rArr e=3#.

#:. b^2=a^2(e^2-1)=1^2(3^2-1)=8#.

# rArr S : x^2/1^2-y^2/8=1, or, 8x^2-y^2=8#.