Answer
$\dfrac{(x-3)^{2}}{16}-\dfrac{(y-2)^{2}}{9}=1$
Work Step by Step
The distance between two vertices is $8$ from the given points.
This means
$\sqrt {(x+2)^2+(y-2)^2}-\sqrt {(x-8)^2+(y-2)^2}=\pm 8$
or, $\sqrt {(x+2)^2+(y-2)^2}=\pm 8+\sqrt {(x-8)^2+(y-2)^2}$
or, $5x-31=\pm4 \sqrt {(x-8)^2+(y-2)^2}$
Hence, $\dfrac{(x-3)^{2}}{16}-\dfrac{(y-2)^{2}}{9}=1$
