Question 1176576
 

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}}} = ±{{{(4/3)x}}}  tell us that h=0 and k=0 
 The center is the point ({{{0}}},{{{0}}}).

Foci is  given to be, (±{{{10}}}, {{{0}}}), tells us that the hyperbola is the horizontal transverse type with the equation in item 1

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

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

The distance to the focus is linear eccentricity {{{c}}}.
The point (10, 0)  shows us the value of {{{c=10}}}:

{{{c^2=a^2+b^2}}}

{{{y }}}= ±{{{(4/3)x}}} shows us that

->{{{b/a=4/3}}}
{{{b=4a/3}}}
{{{10^2=a^2+(4a/3)^2}}}
{{{100=a^2+16a^2/9}}}
{{{100=25a^2/9}}}
{{{900/25=a^2}}}
{{{36=a^2}}}
{{{a=6}}}

then

{{{b/6=4/3}}}
{{{b=24/3}}}
{{{b=8}}}



{{{x^2/6^2-y^2/8^2=1}}}

and your equation in standard form is: 

{{{x^2/36-y^2/64=1}}}


{{{drawing( 600, 600, -15, 15, -15, 15,
circle(10,0,.12),circle(-10,0,.12),
locate(10,0.7,F(10,0)),locate(-10,0.7,F(-10,0)),
 graph( 600, 600, -15, 15, -15, 15,-(4/3)x ,(4/3)x,sqrt(64(-1+x^2/36)), -sqrt(64(-1+x^2/36)))) }}}