How to calculate ellipse sector area *from a focus* 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.

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 
