Answer
$\frac{x^2}{12}-\frac{y^2}{48}=1$
Work Step by Step
Vertices: $(±a,0)=(±2\sqrt 3,0)$
$a=2\sqrt 3$
The hyperbola passes through the point $(4,4)$:
Hyperbola with horizontal transverse axis:
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\frac{4^2}{(2\sqrt 3)^2}-\frac{4^2}{b^2}=1$
$\frac{16}{12}-\frac{16}{b^2}=1$
$\frac{16}{12}-1=\frac{16}{b^2}$
$\frac{4}{12}=\frac{16}{b^2}$
$b^2=48$
$b=4\sqrt 3$
Finally:
$\frac{x^2}{(2\sqrt 3)^2}-\frac{y^2}{(4\sqrt 3)^2}=1$
$\frac{x^2}{12}-\frac{y^2}{48}=1$
