Find the equation of the hyperbola whose foci are $\left(6,4\right)and\left(-4,4\right)$ and eccentricity is 2.

Solution: Let $\left(h,k\right)$ are the coordinates of the center of the hyperbola.
Then, equation of the hyperbola,
$\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$
Now, center of the hyperbola is the midpoint of the foci.
Here, foci are given $\left(6,4\right)\text{and}\left(-4,4\right).$
$\therefore h=\frac{6-4}{2}=1\text{and}k=\frac{4+4}{2}=4$
So, the equation becomes,
$\frac{{\left(x-1\right)}^2}{a^2}-\frac{{\left(y-4\right)}^2}{b^2}=1$
Now, distance between the foci is given as $2ae.$
$\therefore 2ae=\sqrt{{\left(6-\left(-4\right)\right)}^2+{\left(4-4\right)}^2}=10$
It is given that $e=2$.
$\therefore 2a\left(2\right)=10$
$\Rightarrow a=\frac{5}{2}$
$\Rightarrow b^2=a^2\left(e^2-1\right)=\frac{25}{4}\left(4-1\right)=\frac{75}{4}$
So, the equation of the hyperbola is ,
$\frac{{\left(x-1\right)}^2}{\frac{25}{4}}-\frac{{\left(y-4\right)}^2}{\frac{75}{4}}=1$
$\Rightarrow \frac{4{\left(x-1\right)}^2}{25}-\frac{4{\left(y-4\right)}^2}{75}=1$