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I need to find the area of intersection of two concentric ellipses. I imagine the concentricity should make this simpler than the general case, but the ellipses could be rotated.

Alternatively, if there is no simple solution to this. I could live with the area of parallelogram that is formed by the intersections points (when there are 4).

I don't have background in math or geometry, so any pointers is appreciated. Intersecting ellipses:

enter image description here

N. Bar
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Sia Rezaei
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  • Do you have the equations of the ellipses? – N. Bar Aug 03 '19 at 21:24
  • Yes, They are of the form found here: https://math.stackexchange.com/questions/426150/what-is-the-general-equation-of-the-ellipse-that-is-not-in-the-origin-and-rotate/434482#434482 Except h, and k are zero. – Sia Rezaei Aug 03 '19 at 21:37
  • Probably easiest to do this in polar coordinates. – amd Aug 03 '19 at 23:42
  • There's one special case asked here and we may have an oblique case as follows:

    $$\begin{array}{rcl} ax^2+bxy+cy^2<1 & \cap & cx^2-bxy+ay^2<1 \ A &=& \dfrac{8}{\sqrt{4ac-b^2}} \tan^{-1} \sqrt{\dfrac{a+c-\sqrt{(a-c)^2+b^2}}{a+c+\sqrt{(a-c)^2+b^2}}} \ a+c & > & 0 \end{array}$$

     – Ng Chung Tak Aug 04 '19 at 10:10

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This is a partial answer.

I would say the hardest part is likely finding the points of intersection. In this post I discuss the generic approach, which involves solving a quadratic equation of at least degree $3$. However, since your ellipses are both centered around the origin, a quadratic equation will suffice.

Rewrite your ellipses in the following form (by expanding the numerator of the form you say you have):

$$a_1x^2+b_1y^2+c_1xy=d_1\qquad a_2x^2+b_2y^2+c_2cy=d_2$$

Then you can combine them: take $d_2$ times the first equation minus $d_1$ times the second and you get

$$(a_1d_2-a_2d_1)x^2+(b_1d_2-b_2d_1)y^2+(c_1d_2-c_2d_1)xy=0$$

This is the equation of a pair of lines passing through the origin and through the points of intersection. Next compute

$$t = \sqrt{(c_1d_2-c_2d_1)^2-4(a_1d_2-a_2d_1)(b_1d_2-b_2d_1)}$$

The lines are now described by one of these equations:

\begin{align*} 2(a_1d_2-a_2d_1)x + (c_1d_2-c_2d_1\pm t)y &= 0 \\ (c_1d_2-c_2d_1\pm t)x + 2(b_1d_2-b_2d_1)y &= 0 \end{align*}

Generally just picking one of these two equations is enough, since with the $\pm$ in it it will already give you both the lines. If you are unlucky, though, and one of the lines coincides with one of the coordinate axes, then either $(a_1d_2-a_2d_1)$ or $(b_1d_2-b_2d_1)$ will be zero. Using the equation with the zero there will cause one of the equations to become $0x+0y=0$ for one of the cases of the $\pm$. In those cases better use the equation where there is no zero.

Now you have the equation of a line, and you can intersect that with the ellipse. And then you have the points of intersection, and can use these for angle computation. I will not include all details for that in my answer, but I invite other answers to cover these parts.

MvG
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