Answer
Equation of the hyperbola:
${\left( {\frac{{x - 2}}{5}} \right)^2} - {\left( {\frac{y}{{10\sqrt 2 }}} \right)^2} = 1$
Work Step by Step
From the vertices $\left(-3,0\right)$ and $\left(7,0\right)$ we obtain the center of the hyperbola at $C = \left( {\frac{{7 - 3}}{2},0} \right) = \left( {2,0} \right)$.
Since the vertices are $\left(-3,0\right)$ and $\left(7,0\right)$, the distance between the vertices is $10$.
The hyperbola centered at the origin would have equation ${\left( {\frac{x}{a}} \right)^2} - {\left( {\frac{y}{b}} \right)^2} = 1$.
If the center is translated to $C=\left(2,0\right)$, the equation has the form:
(1) ${\ \ \ }$ ${\left( {\frac{{x - 2}}{a}} \right)^2} - {\left( {\frac{y}{b}} \right)^2} = 1$
Notice that $a$ and $b$ are unchanged after translation in this equation. So, we will work with the hyperbola as if it is in the standard position to obtain $a$ and $b$. Then, we substitute the results in equation (1) to obtain the equation of the hyperbola when it is centered at $C=\left(2,0\right)$.
Since the distance between vertices is $10$, in standard position the focal vertices are at $\left( { \pm a,0} \right) = \left( { \pm 5,0} \right)$. By Theorem 4, the eccentricity for a hyperbola is $e = \frac{c}{a}$.
So,
$e = 3 = \frac{c}{5}$, ${\ \ }$ $c=15$
By Theorem 2, we have $c = \sqrt {{a^2} + {b^2}} $. So,
$b = \sqrt {{c^2} - {a^2}} = \sqrt {{{15}^2} - {5^2}} = \sqrt {200} = 10\sqrt 2 $
Substituting $a$ and $b$ in equation (1) gives
${\left( {\frac{{x - 2}}{5}} \right)^2} - {\left( {\frac{y}{{10\sqrt 2 }}} \right)^2} = 1$
