Question 1176575
 

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}}}


Foci is  given to be, ({{{3}}}, {{{0}}}) and ({{{-3}}}, {{{0}}}), tells us that the hyperbola is the horizontal transverse type with the equation in item 1



{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}.... since the center is half way between foci, so it is at origin; so, {{{h=0}}} and {{{k=0}}}


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

The distance to the focus is linear eccentricity {{{c}}}.
The point ({{{3}}}, {{{0}}})  shows us the value of {{{c=3}}}:
The distance to the Vertices is {{{a}}}.
The point ({{{2}}}, {{{0}}}) shows us the value of {{{a=2}}}


{{{c^2=a^2+b^2}}}
{{{3^2=2^2+b^2}}}
{{{3^2-2^2=b^2}}}
{{{9-4=b^2}}}
{{{b^2=5}}}
{{{b=sqrt(5)}}}


then your equation is:

{{{x^2/4-y^2/5=1}}}

{{{drawing( 600, 600, -10, 10, -10, 10,
circle(3,0,.12),circle(-3,0,.12),circle(2,0,.12),circle(-2,0,.12),
locate(3,0.5,F(3,0)),locate(-3,0.5,F(-3,0)),locate(2,0.7,v(2,0)),locate(-2,0.7,v(-2,0)),
 graph( 600, 600, -10, 10, -10, 10, sqrt(5(x^2/4-1)), -sqrt(5(x^2/4-1)))) }}}