PreCalculas help

I really need help
Find the equation of the hyperbola
1.Vertices(-4,0)(4,0)
Asymptotes: y=1/3x, y=-1/3x
2.Vertices(-6,0)(6,0)
Asymptotes: y=4/3x, y=-4/3x

For the first one i got ((x-9)^2+y^2-(x+9)^2+y^2+16=0

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  1. good heavens, whatever did you do to get that equation ?

    I made a sketch, clearly the centre is at (0,0)
    My sketch contains the containing rectangle formed by the asymptotes and the vertices
    so a = 4, and from the asymptotes, b/a = 1/3
    b/4 = 1/3
    b = 4/3

    equation is:
    x^2 /16 - y^2 /(16/9) = 1
    or
    x^2 - 9y^2 = 16

    do the same for the 2nd question

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  2. Thank you

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