Answer
Equation of the ellipse:
${\left( {\frac{{x - 1}}{6}} \right)^2} + {\left( {\frac{y}{{3\sqrt 3 }}} \right)^2} = 1$
Work Step by Step
By Theorem 1, we have $P{F_1} + P{F_2} = 2a$. So, $a=6$.
Since the foci are ${F_1} = \left( {4,0} \right)$ and ${F_2} = \left( { - 2,0} \right)$, the center of the ellipse is at $C = \left( {\frac{{4 - 2}}{2},0} \right) = \left( {1,0} \right)$.
The ellipse 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 $C=\left(1,0\right)$, the equation has the form:
(1) ${\ \ \ }$ ${\left( {\frac{{x - 1}}{a}} \right)^2} + {\left( {\frac{y}{b}} \right)^2} = 1$.
Notice that $a$ and $b$ are unchanged after translation in this equation. So, we will work with the ellipse as if it is in the standard position to obtain $a$ and $b$. Then, we substitute the results in equation (1) to obtain the equation of the ellipse when it is centered at $C=\left(1,0\right)$.
Since the foci are ${F_1} = \left( {4,0} \right)$ and ${F_2} = \left( { - 2,0} \right)$ and the center of the ellipse is at $C=\left(1,0\right)$. In standard position the foci are at $\left( { \pm c,0} \right) = \left( { \pm 3,0} \right)$.
By Theorem 1, we have $c = \sqrt {{a^2} - {b^2}} $. So,
$b = \sqrt {{a^2} - {c^2}} = \sqrt {36 - 9} = \sqrt {27} = 3\sqrt 3 $
Substituting $a$ and $b$ in equation (1) gives
${\left( {\frac{{x - 1}}{6}} \right)^2} + {\left( {\frac{y}{{3\sqrt 3 }}} \right)^2} = 1$
