I know this is a homework but then I need to know how to solve this stuff. Just this one question will do to have a reference to answer the other questions that are like this. Please teach me the process. I know some of you will vote down this question.
So here it is. From the given, find an equation. Foci at $(-1,4)$ and $(7,4)$ and the length of the transverse axis is $\frac{8}{3}$.
Well, you will have to mention the center of the hyperbola in the problem. The best way of getting the hyperbola from the two given foci and the length of transverse axis is to just use the basic graphical definition of the hyperbola, and this works for any given pair of foci and any value for the length of transverse axis.
A hyperbola is the locus of the points such that the difference of distances of that point from two given points, which we call foci, is a fixed-length equal to the length of the transverse axis. So, in your situation the equation of the hyperbola in the crudest form will be as following:
$\sqrt{(x-(-1))^2+(y-4)^2}-\sqrt{(x-7)^2+(y-4)^2}=\pm\frac{8}{3}$.
Now, you can get rid of square roots by squaring both sides and then you will get another square root which you need to isolate and get rid of it by squaring once again. Do not worry. The answer in this particular case will be much simpler. I agree this method will not provide you with the best of the solutions to the problem you asked because your problem has a transverse axis which is parallel to the x-axis, with center the midpoint of the foci which is just $(3,4)$, and the transverse axis $y=4$. So, the answer is just $\frac{(x-3)^2}{(4/3)^2}-\frac{(y-4)^2}{(4/3)^2+4^2}=1$.
Don't forget that the center of the hyperbola is at (3,4) (midpoint of foci). So replace $x^2$ and $y^2$ with $(x-3)^2$ and $(y-4)^2$.
The distance between two foci is given as $2c$. Here distance between the points is $8$ as the $y$ coordinate is constant we do $7-(-1)=8$ therefore $2c=8$ and $c=4$. Then the length of the transverse axis is $\frac{8}{3}$. The length of transverse axis is $2a$ therefore the value of $a=\frac{8}{3}\cdot\frac{1}{2}=\frac{4}{3}$.
We know in hyperbola $c^2-a^2=b^2$ putting the values as $c=4$ and $a=\frac{4}{3}$, $b^2=\frac{128}{9}$, $a^2=\frac{16}{9}$.
Therefore equation of hyperbola is
$$\frac{9x^2}{16}-\frac{9y^2}{128}=1$$
