Multiplication of Points on an Ellipse

What is this about?

8 July 2014, Created with GeoGebra


Previously it has been shown that, if in a cyclic hexagons two pairs of opposite sides are parallel, so are the remaining sides. Besides an Euclidean proof, it was argued that the assertion is a special case of Pascal's theorem and therefore is projective in nature, meaning that it holds for conic sections in general.

If in a hexagon inscribed in a conic two pairs of opposite sides are parallel the same is true for the remaining pair.

Parallel lines meet at a point at infinity. Two pairs of parallel opposite sides of a hexagon determine two points at infinity joint by the line at infinity. According to Pascal's theorem, the three points of intersection of the opposite sides of a hexagon inscribed in a conic are collinear. So if two of them lie at infinity, the same holds for the third intersection. But intersecting at infinity exactly means being parallel.

An application

Given a conic section $\mathbb{c}$ and point $I\in \mathbb{c}.$ For any two points $M$ and $N$ on $\mathbb{c}$ define a third point $M*N$ such that the chord from $I$ to $M*N$ is parallel to $MN.$

Prove that "*" is a binary operation that defines an abelian group on $\mathbb{c}.$

The proof for the case where $\mathbb{c}$ is a circle holds - word for word - in the general case.

An algebraic view

Points on a unit circle in parameterized form are given by pairs $C(t)=(\cos t,\,\sin t),$ $t\in [0,2\pi).$ Points on an ellipse $\displaystyle\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ are parameterized as $E(t)=(a\cos t,\, b\sin t).$ The correspondence between two parameterizations maps an ellipse onto a circle. I wish two show that this mapping leaves parallel chords parallel.

Pick four parameters $u_{1},u_{2},v_{1},v_{2}.$ This is true that $C(u_{1})C(u_{2})\parallel C(v_{1})C(v_{2})$ is equivalent to $E(u_{1})E(u_{2})\parallel E(v_{1})E(v_{2}).$ Indeed, the explicit condition for $E(u_{1})E(u_{2})\parallel E(v_{1})E(v_{2})$ comes from the compariosn of the slopes:

$\displaystyle\frac{b\sin u_{1}-b\sin u_{2}}{a\cos u_{1}-a\cos u_{2}}=\frac{b\sin v_{1}-b\sin v_{2}}{a\cos v_{1}-a\cos v_{2}},$

which is obviously equivalent to the condition for $C(u_{1})C(u_{2})\parallel C(v_{1})C(v_{2}):$

$\displaystyle\frac{\sin u_{1}-\sin u_{2}}{\cos u_{1}-\cos u_{2}}=\frac{\sin v_{1}-\sin v_{2}}{\cos v_{1}-\cos v_{2}}.$

It follows that the abelian groups defined for ellipse above and for circle elsewhere are naturally isomorphic.


The problem stemmed from the one posted at the CutTheKnotMath facebook page by Leo Giugiuc (Romania). Leo, posted the problem for an ellipse and supplied an elegant algebraic solution. The idea of using Pascals' theorem is due to Telv Cohl.

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