Question

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

Answer 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

Answer 2

Thank you