Question 1138308

 Find an equation in standard form for the hyperbola with vertices at ({{{0}}}, ±{{{4}}}) 

and asymptotes at {{{ y}}} = ± {{{(1/4) x}}} 


standard form:

since the vertices are at (0, ±4), that means the parabola opens up and down, and {{{a = 4}}} (the distance from center to vertex)

the center will be halfway between the vertices at:({{{h}}}, {{{k}}}) = ({{{0}}}, {{{0}}}) 

the hyperbola will look like:

 {{{(y-k)^2 / a^2 - (x-h)^2 / b^2 = 1    }}}

 {{{(y-0)^2 / 4^2 - (x-0)^2 / b^2 = 1    }}}

 {{{y^2 / 16 - x^2 / b^2 = 1    }}}

since given asymptotes at  {{{y}}} = ± {{{(1/4) x }}}, the slope of the vertices will be ± {{{a/b}}}, and we have

± {{{1/4}}} = ±{{{ 4/b }}}=> {{{b = 16}}}

substituting in  {{{b = 16}}} : 

 {{{y^2 / 16 - x^2 / 16^2 = 1    }}}

{{{y^2 / 16 - x^2 / 256= 1 }}}