How would I get the area of the portion of the ellipse identified as "back" (to the left of the focus "S") in the drawing and the portion of the ellipse identified as forward (to the right of the focus "S") in the drawing.
EDIT: Image of the ellipse has been changed to better coincide with the answer given by @of course.
Applying the idea in my comment above, and the notation depicted in your figure, we want to horizontally scale the ellipse, and turn it into a circle of radius $R$.
Define the image of scaling as follows:
$ x' = \sin(\beta) \ x $
$ y' = y $
Since the original equation of the ellipse is
$ \dfrac{ x^2 \sin^2( \beta) }{R^2} + \dfrac{y^2}{R^2} = 1 $
Then after applying the scaling, it becomes the circle
$ x'^2 + y'^2 = R^2 $
Originally the line separating the two parts of the ellipse is located at $x = s $ where $ - \dfrac{R}{\sin(\beta)} \le s \le \dfrac{R}{\sin( \beta)} $
The corresponding line for the scaled ellipse is
$ x' = s' $ where $ s' = s \sin(\beta) $
Now the area of the left part of the circle to the left of this vertical line is
$ A'_{Left} = \dfrac{1}{2} R^2 \phi - \dfrac{1}{2} R^2 \sin( \phi) $
where $ \phi = 2 \cos^{-1} \left( \dfrac{ -s'} {R} \right) = 2 \cos^{-1} \left( - \dfrac{ s \sin(\beta)}{R} \right) $
The other part is
$ A'_{Right} = \pi R^2 - A'_{Left}$
The corresponding areas in the original ellipse are found simply by scaling $A'_{Left}$ and $A'_{Right} $ by the reciprocal of the scaling factor that we used to get the circle from the ellipse, (and this scaling factor is $ \sin \beta $). Therefore, the left and right areas of the ellipse are
$ A_{Left} = \left( \dfrac{1}{\sin \beta} \right) A'_{Left} $
$ A_{Right} = \left( \dfrac{1}{\sin \beta} \right) A'_{Right} $