How do you calculate the area of a sector of an ellipse when the angle of the sector is drawn from one of the focii? In other words, how to find the area swept out by the true anomaly?
There are some answers on the internet for when the sector is drawn from the center of the ellipse, but not from the focii.
Ted's answer already outlined what I'd do, but left a lot of things for readers to work out or look up. So here I'll give details on these. The main idea is to transform the problem to the circle, where areas are computed more easily.
The polar parametrization of an (axis-aligned) ellipse from its focus is given by
$$\overline{FP}=r(\theta)=\frac{a(1-e^2)}{1+e\cos\theta}$$
where $a$ is the semi-major axis, $e$ is the eccentricity and $\theta$ is the angle, i.e. the true anomaly. Spelled out in coordinates, this is
$$\overrightarrow{FP}= \begin{pmatrix}r(\theta)\cos\theta\\r(\theta)\sin\theta\end{pmatrix}$$
Now stretch the whole things in the direction of its minor axis (i.e. its $y$ direction) by a factor of
$$\frac ab=\frac1{\sqrt{1-e^2}}$$
The result will be a circle of radius $a$. The origin is still the point that used to be the focus. The distance between that point and the center is $\overline{CF}=ae$ (also called linear eccentricity or focal distance). So add that to $x$ coordinates to move the origin into the center. Together you now have
$$ \overrightarrow{CQ}= \begin{pmatrix}r(\theta)\cos\theta+ae\\ \frac ab r(\theta)\sin\theta\end{pmatrix} = \frac{a}{1+e\cos\theta} \begin{pmatrix}(1-e^2)\cos\theta+e+e^2\cos\theta\\ \sqrt{1-e^2}\sin\theta\end{pmatrix} $$
The angle for that point (against the horizontal axis) is the eccentric anomaly $E$, so it satisfies the following relation:
$$\tan E=\frac{\sqrt{1-e^2}\sin\theta} {(1-e^2)\cos\theta+e+e^2\cos\theta}$$
Using the Weierstrass substitution André mentioned in his answer, you can turn this into
\begin{align*} \frac{2\tan\frac E2}{1-\tan^2\frac E2}&= \frac{\sqrt{1-e^2}2\tan\frac\theta2} {(1-e^2)(1-\tan^2\frac\theta2)+e(1+\tan^2\frac\theta2)+e^2(1-\tan^2\frac\theta2)} \\&= \frac{2\sqrt{(1+e)(1-e)}\tan\frac\theta2}{(1+e)-(1-e)\tan^2\frac\theta2} =\frac{2\sqrt{\frac{1-e}{1+e}}\tan\frac\theta2} {1-\frac{1-e}{1+e}\tan^2\frac\theta2} \end{align*}
By simply comparing both sides you can see
$$\tan\frac E2 = \sqrt{\frac{1-e}{1+e}}\tan\frac\theta2$$
This agrees with the relation Wikipedia mentions. And dealing with the tangent of half the angle has the benefit that you can compute the arctan without worrying about the quadrant. You can simply compute
$$E=2\arctan\left(\sqrt{\frac{1-e}{1+e}}\tan\frac\theta2\right)$$
except for the special case when $\theta=\pm\pi$ which yields an infinite tangent but results in $E=\pm\pi$ as well.
Now you know that the area of a circular sector is proportional to the angle, so for a circle of radius $a$ this would be $\frac12a^2E$. That's the yellow and cyan areas taken together, i.e. the sector $ACQ$.
But that area is subtended by the center, not by the focus. Therefore subtract a triangle of base $ae$ and height $a\sin E$ (the cyan triangle $\triangle FCQ$) and you get the area subtended by the focus ($AFQ$ colored yellow):
$$ \tfrac12a^2E-\tfrac12a^2e\sin E=\tfrac12a^2\left(E - e\sin E\right) $$
This is still an area in the circle. To get back to the original ellipse, you have to undo the scaling, i.e. scale the $y$ direction by $\frac ba$. This scales areas by the same factor. Thus the area you were asking about ($AFP$ tinted red) would be
$$\tfrac12ab\left(E-e\sin E\right)$$
As you vary $\theta$ from $0$ to $2\pi$, the above area will vary from $0$ to $ab\pi$. Since equal areas are swept in equal time, this is closely related to the mean anomaly which sweeps from $0$ to $2\pi$ in constant time.
$$M=E-e\sin E$$
This result is again mentioned on Wikipedia, with a reference to Kepler's equation. So in terms of the mean anomaly, you'd get the area as
$$\tfrac12abM$$
just as ben wrote in this comment.
The area will be $$\int_{\theta_1}^{\theta_2}\frac{1}{2}r^2d\theta,$$ where $r=r(\theta)$ is the equation of the ellipse, with polar origin at the focus.
Imagine an ellipse with semi-major axis $a$ and eccentricity $e$, and with one of the foci at the origin, and the other focus on the half-line $\theta=0$ (so to the "right" of the origin). Then the ellipse has polar equation $$r=\frac{a(1-e^2)}{1-e\cos\theta}.$$
The rest is unpleasant. There is a closed form for the antiderivative. One might ask Wolfram Alpha. If you want to do it yourself, you can use the Weierstrass substitution. But there is undoubtedly a neater way.
Ben, here's a better suggestion. You can stretch a circle to make an ellipse and, if you start with a unit circle, area is magnified by the factor of $ab$, where $a$ and $b$ are the semi-axes, as usual. Take a point at $(-R,0)$ inside the unit circle and consider the sector it subtends to $(1,0)$ and $(\cos t, \sin t)$. You can find the area pretty easily: I get $\frac 12(t+R\sin t)$. Now stretch by the fudge factor and figure out how to match up $R$ with your focus and $t$ with your arbitrary point on the ellipse.
MY APOLOGIES. THIS ANSWER IS COMPLETELY WRONG. I HAVE NO IDEA WHY I THOUGHT THE CENTRAL AREA OF AN ELLIPSE WAS theta times a times b, BUT IT IS NOT. I AM LEAVING THIS ANSWER HERE FOR ARCHIVAL PURPOSES, BUT IT IS INACCURATE.
The area swept out from the focus of an ellipse is:
$ b \left(a \theta -b \sin (\theta ) \sqrt{\frac{(a-b) (a+b)}{a^2 \cos ^2(\theta )+b^2 \sin ^2(\theta )}}\right) $
where $\theta$ is the central angle, $a$ is the semimajor axis, and $b$ is the semiminor axis.
This is actually a simplified version of a portion of @MvG's answer, and a bit of a cheat, since you normally wouldn't have the central angle, but I believe the derivation (below) might be useful to some.
The area swept out from the center of an ellipse is $a b \theta$ where $\theta$ is the central angle, $a$ is the semimajor axis, and $b$ is the semiminor axis:
To find the area from the focus, we simply subtract off this purple triangle, where F is the focus in question:
The distance from the center of an ellipse to either focus is $\sqrt{a^2-b^2}$ giving us the base of this triangle.
To find the height, we start by knowing two things about $(x,y)$:
$\{x=a \cos (t),y=b \sin (t)\}$
for some value of $t$. Note that $t\neq \theta$.
$\frac{y}{x}=\tan (\theta )$
Combining the two equations, we have:
$\frac{b \sin (t)}{a \cos (t)}=\tan (\theta )$
Solving for t:
$t=\tan ^{-1}(a \cos (\theta ),b \sin (\theta ))$
Note that we must use the two argument form of $tan ^{-1} ()$ to make sure $t$ is in the correct quadrant.
Plugging back in for $x$ and $y$ and applying trignometric identities and other simplifications:
$x=\frac{a^2 \cos (\theta )}{\sqrt{a^2 \cos ^2(\theta )+b^2 \sin ^2(\theta )}}$
$y=\frac{b^2 \sin (\theta )}{\sqrt{a^2 \cos ^2(\theta )+b^2 \sin ^2(\theta )}}$
We now have the height of the triangle, $y$, as above. Apply the area formula gives us:
$b^2 \sin (\theta ) \sqrt{\frac{(a-b) (a+b)}{a^2 \cos ^2(\theta )+b^2 \sin ^2(\theta )}}$
Subtracting that from the original $a b \theta$, we get:
$ b \left(a \theta -b \sin (\theta ) \sqrt{\frac{(a-b) (a+b)}{a^2 \cos ^2(\theta )+b^2 \sin ^2(\theta )}}\right) $
Following the approach I used for Evaluating $\int_a^b \frac12 r^2\ \mathrm d\theta$ to find the area of an ellipse, parametrize the ellipse as:
$x(t)=a \cos (t)$
$y(t)=b \sin (t)$
with $a>b$ (the case $a<b$ is symmetric). Note that $t$ is neither the central angle nor the focal angle. The foci are then on the x-axis with $x=\pm\sqrt{a^2-b^2}$ and the top right quadrant of the ellipse looks like this:
where $d = a \cos (t)-\sqrt{a^2-b^2}$
This gives us:
$\tan (\theta ) = \frac{b \sin (t)}{a \cos (t)-\sqrt{a^2-b^2}}$
Solving for t, we have:
$t(\theta) = \cos ^{-1}\left(\frac{a \sqrt{(a-b) (a+b)} \tan ^2(\theta )+b^2 \sec (\theta )}{a^2 \tan ^2(\theta )+b^2}\right)$
provided that $\theta<\pi/2$ (the case $\theta>=\pi/2$ is similar, though not symmetric, and is left as an exercise for the reader)
We can now reparametrize the ellipse using the focal angle:
$x(\theta )=a \cos (t(\theta ))$
$y(\theta )=b \sin (t(\theta ))$
or:
$x(\theta) = \frac{a \left(a \sqrt{(a-b) (a+b)} \tan ^2(\theta )+b^2 \sec (\theta )\right)}{a^2 \tan ^2(\theta )+b^2}$
$y(\theta) = b \sqrt{1-\frac{\left(a \sqrt{(a-b) (a+b)} \tan ^2(\theta )+b^2 \sec (\theta )\right)^2}{\left(a^2 \tan ^2(\theta )+b^2\right)^2}}$
and compute the radius squared for a given $\theta$ (we could also compute the radius itself, but we won't need it):
$r(\theta )^2=x(\theta )^2+y(\theta )^2$
Substituting and simplifying, this gives us:
$r(\theta)^2 = \frac{a^2 \tan ^2(\theta ) \left(\left(a^4-a^2 b^2+b^4\right) \tan ^2(\theta )+2 b^4\right)+b^4 (a-b) (a+b) \sec ^2(\theta )+2 a b^2 ((a-b) (a+b))^{3/2} \tan ^2(\theta ) \sec (\theta )+b^6}{\left(a^2 \tan ^2(\theta )+b^2\right)^2}$
Now we simply integrate $r(\theta)^2/2$ over $d\theta$:
$A(\theta) = \int_0^{\theta } \frac{r(\phi)^2}{2} \, d\phi$
After integrating and simplifying, we have $\text{A}(\theta)=$
$\frac{1}{2} \left(\theta \left(a^2+b^2\right)+b \left(a \cot ^{-1}\left(\frac{b \csc (\theta )}{\sqrt{a^2-b^2}}\right)-\frac{2 b \sin (\theta ) \left(\left(b^2-a^2\right) \cos (\theta )+a \sqrt{(a-b) (a+b)}\right)+a \left(\left(b^2-a^2\right) \cos (2 \theta )+a^2+b^2\right) \tan ^{-1}\left(\frac{a \tan (\theta )}{b}\right)}{\left(b^2-a^2\right) \cos (2 \theta )+a^2+b^2}\right)\right)$
As a reminder, this only works when $a>b$ (though the other case is symmetric) and $\theta<\pi/2$ (the other case is similar, but not symmetric).
Details on how I worked this out: https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-ellipse-focus.m
Newtonian dynamical formulation is easiest if you want area with respect to time because, as per Kepler's second Law equal sectorial areas are swept out in equal intervals of time. Moreover, you gain the original Newtonian insight in this exercise.
You need to integrate the differential equation and convert constant angular momentum $ h= (d\theta/dt) r^2, GM $, latus rectum $p$ etc. into geometrical parameters to feed to the ellipse polar form:
$$r=\frac{a(1-e^2)}{1-e\cos\theta} =\frac p {1-e\cos\theta}. $$
The derivation starting with inverse square law radial/circumferential equilibrium is not that difficult either.
Converting the way Newton derived it first from dynamical situation to geometrical / polar form above is now standard/canonical.He saw no difference between the dynamical and geometric sides of planetary motion.
I have the ($ r-\theta $) ODE geometric derivation for conics also, but I advise to do dynamics derivation at first.
