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.
On
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.
On
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
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.