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November 29

Generalizing a Property of the Ellipse

The ellipse has the property that the normal (perpendicular) to the tangent bisects the angle formed by the point of tangency with the two foci. What is the locus of the points in a plane with the property that the normal (perpendicular) to the tangent trisects the angle formed by the point of tangency with two fixed points ? — 79.113.228.82 (talk) 05:38, 29 November 2015 (UTC)[reply]

Since no one has responded in more than 24 hours I'll do so even though I don't really have an answer. First, it's a bit nit picky but the word "locus" usually means the set of points satisfying a condition, but the condition you're placing on the curve depends on the curve itself. So it's not a locus in the traditional sense. The condition given is more like a differential equation so the result is a family of curves rather than a single curve as would be defined as a locus. As I understand it, the angle bisection property of an ellipse is a consequence of the fact the sum of the distances to the two foci is a constant. You get the same thing with a hyperbola where the difference of the distances to the two foci is a constant, only this time the lines are bisected the other way. With a parabola you get the tangent bisects the line to the focus and a line parallel to the axis, which you might call the line to the focus at infinity. So you might look at replacing the condition that the sum of the distances to the foci are constant with one where the distance to the first focus plus twice the distance to the second one is constant. This defines a member of a family of curves known Cartesian ovals. Unfortunately, this doesn't lead to a simple relationship between the angles, though you do get a relationship between their cosines as in Snell's law. To get back to the original question, it's much easier to find a differential equation for a family of curves than to find a family of curves that satisfy a differential equation. When I tried to set this up as a differential equation, even simplifying by placing one of the foci at infinity as in the parabola, the result looked too complicated to be solved easily and there's no guarantee that it would have a solution in closed form. There may be a trick I'm overlooking though so I won't claim there is no such solution. --RDBury (talk) 16:48, 30 November 2015 (UTC)[reply]

Actually there was a trick I was missing, at least to find a solution when one of the foci is at infinity, namely polar coordinates. In fact it works for angle bisection, trisection, tetrasection, or any other linear relationship. The angle of inclination of the line from the point to the origin (taken to be one focus) is θ. The bisector of this line and a horizontal line will have inclination θ/2, the trisector angle θ/3 or 2θ/3, or generally, say kθ. So we want the angle of inclination of the tangent, the tangential angle φ, to be kθ. The angle between the tangent and the line to the origin is called the polar tangential angle ψ and the relationship between these angles is φ = θ + ψ. Eliminating φ we get ψ=(k-1)θ or ψ=nθ where n=k-1. Now ψ is given by (see the tangential angle article)

\tan \psi ={\frac {r}{dr/d\theta }}

.

So we get the differential equation

\tan n\theta ={\frac {r}{dr/d\theta }}

.

This is easily solved by separation of variables to get

r^{n}=a^{n}\sin(n\theta )

which is the equation for a sinusoidal spiral. For k=1/2 (bisection), n=-1/2 this is the parabola as expected. For k=1/3 or 2/3 (trisection), n=-2/3 or -1/3 which is Tschirnhausen cubic in the second case and an unnamed curve in the first. I don't know if this can be adapted to the case where both foci are finite, but it now seems slightly less improbable that a solution can be found. --RDBury (talk) 18:13, 30 November 2015 (UTC)[reply]

I don't understand what angle exactly the normal to the tangent is supposed to trisect... — 82.137.52.216 (talk) 01:24, 1 December 2015 (UTC)[reply]

The plot at Tschirnhausen cubic isn't a good one to look at because it has been shifted and stretched. If instead you look at the plot at http://mathworld.wolfram.com/TschirnhausenCubic.html then the angle to look at is focus-curve-focus, where the first focus is where the axes meet, and the second focus is infinitely far towards the left. Egnau (talk) 18:05, 1 December 2015 (UTC)[reply]

Okay, I think I finally got it: You mean a point situated at negative infinity, or

(-\infty ,0),

which, when united with the point of tangency, yields basically a line parallel to the horizontal axis, passing through the point of tangency, and the normal to the tangent trisects the angle formed by the position vector with the aforementioned line. — 79.113.227.158 (talk) 03:53, 2 December 2015 (UTC)[reply]

Putting two foci at (0.1, 0.0) and (0.9, 0.0), I wrote a simple ODE that is valid as long as the curve stays above the x-axis. Wolfram Alpha can solve it numerically and give a rough plot:

- bisects (ellipse) [1],

- trisects with the small angle on the right [2],

- trisects with the small angle on the left [3].

The last case shows that the curve can cross the line between the two foci at a ~~60°~~ 30° angle, which is not something that can happen with an ellipse. Egnau (talk) 19:27, 30 November 2015 (UTC)[reply]

Okay, let's narrow it down a bit: Could you please plot the case when the two "foci" are situated at (±1, 0), and the "generalized ellipse" passes through (0, 1) ? (I think I have an idea of how to modify the parameters from the last two links, but I am not 100% sure). — 79.113.227.158 (talk) 03:53, 2 December 2015 (UTC)[reply]

The modified two links would be [4] and [5]. Unfortunately, with these queries Wolfram Alpha plots only x ≥ 0 and without axes. Removing the condition f(0) = 1 enables the "Sample solution family" section which shows axes, but gives no control over the plotted range. Maybe paying for the "Pro" version would give more control, I don't know. Alternatively there's software for ODE solving. Egnau (talk) 05:37, 2 December 2015 (UTC)[reply]

Resolved

– — Thank you both for your help ! :-)