The way that you'd do this in practice would be as follows: 1. Write down both ellipses as degree-2 polynomials over X, Y -- i.e. in the form A X^2 + 2 B X Y + C Y^2 + 2 D X + 2 E Y + F = 0. We can homogeneise them by adding an extra indeterminate, Z, making them into projective equations of the form: A X^2 + 2 B X Y + C Y^2 + 2 D X Z + 2 E Y Z + F Z^2 2. Find a nonzero linear combination of the two polynomials, such that the discriminant (determinant of [[A B D] [B C E] [D E F]]) is zero; this is a degenerate conic. 3. Take an arbitrary linear factor P X + Q Y + R Z of the degenerate conic; this defines a line L with equation P X + Q Y + R Z = 0. 4. Simultaneously solve this linear equation with the quadratic equation corresponding to the first ellipse. There are two nonreal roots (in the complex projective plane), which must be complex conjugates since the coefficients of the equation are real. 5. Find a linear transformation with real coefficients which maps the coordinates (X, Y, Z) of one of those intersection points to the 'circular point' (1, i, 0). The other intersection point is necessarily mapped to the other circular point, (1, -i, 0). 6. This linear transformation of R^3, viewed as a projective transformation of RP^2, is exactly what you need to convert the original pair of ellipses into circles. Best wishes, Adam P. Goucher
Sent: Sunday, May 26, 2019 at 3:58 PM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Poncelet's porism overpush?
<< we take a (complex conjugate!) pair of intersection points, ... >>
Um --- interactive demo for dummies, anybody? WFL
On 5/26/19, Adam P. Goucher <apgoucher@gmx.com> wrote:
Every pair of cleanly nested ellipses can be converted to a pair of circles under a suitable projective transformation. Specifically, we take a (complex conjugate!) pair of intersection points, and map them to the two circular points at infinity using a projective transformation with real coefficients:
https://en.wikipedia.org/wiki/Circular_points_at_infinity
So both of the questions:
*Arbitrary* (cleanly nested) ellipses? Don't they both need to be circles under the same projection?
are true, because they're equivalent.
Best wishes,
Adam P. Goucher
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