Answer
$\frac{(x-4)^2}{4}-\frac{(y+2)^2}{5}=1$
Work Step by Step
Step 1. From the given conditions, the hyperbola has a center at $(4,-2)$ and a horizontal transverse axis; thus the equation is in the form of
$\frac{(x-4)^2}{a^2}-\frac{(y+2)^2}{b^2}=1$
Step 2. From the given coordinates, we have
$a=6-4=2, c=7-4=3$
Thus
$b^2=c^2-a^2=5$
Step 3. The equation is
$\frac{(x-4)^2}{4}-\frac{(y+2)^2}{5}=1$
