Question 717075
The equation of a hyperbola with a horizontal transverse axis can be written in the form
{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}
I had to look it up (not good with names), but that is what is called the standard form.
(h,k) is the center
Center, vertices, and foci are on the horizontal line {{{y=k}}}
For the vertices,
{{{(x-h)^2/a^2=1}}} --> {{{(x-h)^2=a^2}}} --> {{{abs(x-h)=a}}}
They are at distance {{{a}}} from the center, on line {{{y=k}}}
and {{{2a}}} is the distance between the vertices.
The segment (and the distance) between the vertices is called the transverse axis.
 
So far we know
{{{h=4}}}, {{{k=1}}}, and {{{2a=12}}} --> {{{a=6}}}
 
The eccentricity {{{e}}} is defined based of the focal distance {{{c}}}
(distance from each focus to the center) as
{{{e=c/a}}}
and it turns out that {{{a}}}, {{{b}}} and {{{c}}} are related by {{{c^2=a^2+b^2}}}
We know {{{e=4/3}}} so
{{{4/3=c/6}}} --> {{{c=6*4/3}}} --> {{{c=8}}}
Then, plugging that (along with {{{a=6}}}) into {{{c^2=a^2+b^2}}} we get
{{{8^2=6^2+b^2}}} --> {{{64=36+b^2}}} --> {{{b^2=64-36}}} --> {{{b^2=28}}}
 
Finally, plugging the values given (or very easily found) for {{{a}}}, {{{h}}}, and {{{k}}}, plus the hard earned value for {{{b^2}}} into the standard form we get
{{{highlight((x-4)^2/36-(y-1)^2/28=1)}}}