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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.

enter image description here

rdemo
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    The simplest way is to compress the ellipse and turn it into a circle. Then find the area of the circular cap (which is relatively easy), then multiply this area by the inverse of the scale factor that you applied in the first place. – Hosam Hajeer Mar 02 '24 at 01:39
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    I might turn the equation into a function just above the x-axis, integrate just the half of the elliptical segment above the x-axis, then double to calculate the are of the elliptical segment to the left of the vertical line passing though point s. – Nate Mar 02 '24 at 03:08

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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} $

Hosam Hajeer
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  • Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed. – Xander Henderson Mar 08 '24 at 12:33