Can an ellipse be divided into sectors of equal area? Is there any generalized formula to do so? The total area of the ellipse $A$ has to be divided into $n$ equal sectors such that $A=A_1+A_2+ ⋯+A_n$ given its semi-major axis $a$ and semi-minor axis $b$. Any solution or resources will be of great help?
Dividing an Ellipse into Equal Areas – Geometry
geometry
Related Solutions
I would resize the circles by a factor $k$ (in the figure $k=2$) so that the maximum number of circles is tangent to the ellipse and among them. To determine their centers I would try to solve this complex packing problem by trial and error, starting with the biggest circle (corresponding to the light blue one in the figure) and proceeding with the smaller ones (corresponding to the dark blue and the two equal sized black ones). The original circles would have the same centers.
This is easily solvable with resultants. Just set up the parameters for the coordinates of the foci and c, then write down five pretty obvious polynomial equations, then use the Dixon resultant. Here's the answer (assuming I didn't make a dumb mistake):
$A = \pi*c*s/4$
where
$s = \sqrt{ c^2 - (f_2y-f_1y)^2 - (f_2x-f_1x)^2 }$
The foci are $(f_1x, f_1y)$ and $(f_2x, f_2y)$.
Note that this reduces to the right thing if the foci are equal, or if c = the distance between the foci.
Best Answer
Recall that any ellipse can be obtained by affinity from a circle and that the affinity preserves the ratio between the areas.
Thus we can consider the division into sectors of equal area for a circle and then the corresponding affine transformation to the ellipse.
Notably. as an example, let consider the affinity by $X=ax$ and $Y=by$ such that
$$x^2+y^2=1 \to \frac{X^2}{a^2}+\frac{Y^2}{b^2}=1$$
with reference to the first quadrant, the equations for the lines dividing the circle into $n$ sectors of equal area $A=\frac{\pi}{4n}$ are
$$y=(\tan \theta_k)\cdot x, \quad \theta_k=\frac{k\pi}{2n},\quad k=1,...,n-1$$
the equations for the lines dividing the ellipse into $n$ sectors of equal area $A=\frac{\pi ab}{4n}$ are
$$Y=\frac b a (\tan \theta_k)\cdot X, \quad \theta_k=\frac{k\pi}{2n},\quad k=1,...,n-1$$