Answer
$\frac{(x-3)^{2}}{16}-\frac{(y-2)^{2}}{9}=1$
Work Step by Step
From the given points,
The distance between the two vertices is $8$.
Therefore, the equation of the parabola is:
$\sqrt {(x+2)^2+(y-2)^2}-\sqrt {(x-8)^2+(y-2)^2}=\pm 8$
$\sqrt {(x+2)^2+(y-2)^2}=\pm 8+\sqrt {(x-8)^2+(y-2)^2}$
Squaring both side, we get
$5x-31=\pm4 \sqrt {(x-8)^2+(y-2)^2}$
Thus,
$\frac{(x-3)^{2}}{16}-\frac{(y-2)^{2}}{9}=1$
