Answer
Equation of the conic section:
${\left( {x - 6} \right)^2} - {\left( {\frac{y}{{\sqrt {15} }}} \right)^2} = 1$
Work Step by Step
Since the foci are $\left( {2,0} \right)$ and $\left( {10,0} \right)$, the center of the hyperbola is $C = \left( {\frac{{10 + 2}}{2},0} \right) = \left( {6,0} \right)$.
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 $C = \left( {6,0} \right)$, the equation has the form:
(1) ${\ \ \ }$ ${\left( {\frac{{x - 6}}{a}} \right)^2} - {\left( {\frac{y}{b}} \right)^2} = 1$.
Notice that $a$ and $b$ are unchanged after translation in this equation. So, we will work with the ellipse as if it is in 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 $C = \left( {6,0} \right)$.
Since the foci are $\left( {2,0} \right)$ and $\left( {10,0} \right)$, in standard position the foci would be
$\left( {2,0} \right) - \left( {6,0} \right) = \left( { - 4,0} \right)$ ${\ \ }$ and ${\ \ }$ $\left( {10,0} \right) - \left( {6,0} \right) = \left( {4,0} \right)$
Thus, $\left( { \pm c,0} \right) = \left( { \pm 4,0} \right)$. So, $c=4$.
By Theorem 4 of Section 12.5: $e = \frac{c}{a}$. So,
$a = \frac{c}{e} = 1$
By Theorem 2 of Section 12.5: $c = \sqrt {{a^2} + {b^2}} $, so
$b = \sqrt {{c^2} - {a^2}} = \sqrt {16 - 1} = \sqrt {15} $
Substituting $a$ and $b$ in equation (1) gives
${\left( {\frac{{x - 6}}{1}} \right)^2} - {\left( {\frac{y}{{\sqrt {15} }}} \right)^2} = 1$
${\left( {x - 6} \right)^2} - {\left( {\frac{y}{{\sqrt {15} }}} \right)^2} = 1$
