Dividing an ellipse into equal area

Question

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?

Reminds me of that one Kepler's law: A planet in an elliptic orbit sweeps out equal areas in equal times. — G Tony Jacobs, May 25 '18 at 21:49
What is the origin of the sectors? One of the foci, or the center of the ellipse? — Arthur, May 25 '18 at 22:17
I think that this is (potentially) an interesting question, but I feel like I am forced to vote to put it on hold as "lacking context." If you were to edit your question to to provide an answer to the issue raised by Arthur, I would be more than willing to retract that vote. It might also be worthwhile to spend some time thinking about Kepler's law; a unit speed parameterization of the ellipse would likely give the right answer. — Xander Henderson, May 26 '18 at 03:51
@Arthur: The origin of the sectors lies at the center of the ellipse. — pretty, May 26 '18 at 07:46

user · Accepted Answer · 2018-05-26T08:51:33.463

7

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

$$Y=\frac b a (\tan \theta_k)\cdot X, \quad \theta_k=\frac{k\pi}{2n},\quad k=1,...,n-1$$

edited May 26 '18 at 08:51

answered May 25 '18 at 22:29

user

154,566

Thanks for the information and it worked. – pretty May 26 '18 at 18:09
Well done! You are welcome! Bye – user May 26 '18 at 18:30
I have one more question. Does this solution can also be scaled to superellipse. https://math.stackexchange.com/questions/2797052/dividing-concentric-super-ellipses-to-equal-area-slices/2797062#2797062 – pretty May 26 '18 at 18:33
Yes of course, I’m going to add also something on that if you loke. – user May 26 '18 at 18:40
That was so neat ! Works when first sector radius does not coincide with x-axis? – Narasimham May 27 '18 at 00:19
@Narasimham Yes, it works for any configuration! – user May 27 '18 at 02:06
@gimusi yes please, it would be of great help. – pretty May 27 '18 at 06:48
@pretty Sorry now i can't but I will give a look later! Bye – user May 27 '18 at 07:01
Comment was.. as the sketch did not contain ellipses with inclined axes. – Narasimham May 28 '18 at 10:09

Dividing an ellipse into equal area

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