Question 1176577
 Find the standard form of the equation of the hyperbola with the given characteristics.

Vertices: ({{{0}}}, ±{{{9}}}); asymptotes: {{{y }}}= ±{{{3x}}}

There are two standard forms:

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

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


The point-slope form for the equations of the asymptotes is:

{{{y}}}= ±{{{ m(x-h)+k}}}

Therefore, the equation, {{{y }}}}= ±{{{3x}}}  tell us that {{{h=0}}} and {{{k=0}}}
 
 The center is the point ({{{0}}},{{{0}}}).

the vertices given to be, ({{{0}}}, ±{{{9}}}), tells us that the hyperbola is the {{{vertical}}} transverse type with the equation in item 2

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

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

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


The point ({{{0}}}, {{{9}}}) allows us to discover the value of "{{{a}}}":

{{{9^2/a^2-0^2/b^2=1}}}

{{{81/a^2=1}}}

{{{a^2=81}}}

{{{a=9}}}

the asymptotes allows us to discover the value of "{{{b}}}": 

{{{y=3x}}}, so {{{b=3}}}

Substitute the value of "{{{a}}}"  and “{{{b}}}” into equation:

{{{y^2/9^2-x^2/3^2=1}}}

and your equation in standard form is: 

{{{y^2/81-x^2/9=1}}}


{{{drawing( 600, 600, -10, 10, -20, 20,
circle(0,9,.12),circle(0,-9,.12),locate(0,9,v(0,9)),locate(0,-9,v(0,-9)),
 graph( 600, 600, -10, 10, -20, 20,3x,-3x,- sqrt(81(1+x^2/9)),sqrt(81(1+x^2/9))) )}}}