From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: Parallel curve to ellipse Date: 20 Mar 2000 23:55:49 GMT Newsgroups: sci.math,geometry.research,geometry.college Summary: [missing] In article <38d628e8.0@news.smith.edu>, Joseph O'Rourke wrote: >I would like to find a clean representation, say as an implicit >equation, or a nice parametrization, for a curve parallel to an >ellipse, i.e., the curve produced by displacing inward a fixed >distance orthogonal to the tangent at each point of an ellipse. No problem. Translate, rotate, and scale so that you're trying to move one unit inward from the ellipse (x/a)^2 + (y/b)^2 = 1. That's easy. You can walk to every point on the ellipse with the parameterization p(t) = (a cos(t), b sin(t)). A tangent vector there is v(t) = p'(t) = (-a sin(t), b cos(t)), and a normal vector is then n(t) = (b cos(t), a sin(t)). Assuming a, b > 0, this points outward. So you just want to look at the points p(t) - n(t)/||n(t)|| = ( (a - b/q) cos(t), (b - a/q) sin(t) ), where q(t) = sqrt( b^2 cos(t)^2 + a^2 sin(t)^2 ). Tada, that's a parameterization. This parameterization exhibits this as a closed curve (the parameterization is periodic). Also useful is a parameterization almost by rational functions. It's not as commonly known that as you range over all values for t, the pairs (cos(t), sin(t)) will be essentially the same as the pairs ((1-u^2)/(1+u^2), (2u)/(1+u^2)) as u ranges over the real line. You can explain this geometrically but I don't wish to do so here; just note that we then have the parameterization x = (a - b/q) (1-u^2)/(1+u^2) y = (b - a/q) (2u)/(1+u^2) where q^2 = b^2 ((1-u^2)/(1+u^2))^2 + a^2 ((2u)/(1+u^2))^2, so that x and y are given parametrically as rational functions of u and q (which involves the square root of a quartic polynomial in u ). So now you can describe the curve as the set of points (x,y) for which real numbers (u,q) exist (with q>0) making these three equations true. You can clear denominators to turn this into a set of three polynomial equations in x, y, u, q. You can turn this description into a single implicit equation -- and I suppose I ought to remind you that there is essentially only one such , that is, the ideal of polynomial functions vanishing on this curve is principal -- by using elimination techniques. But let me warn you in advance that this is hardly what I would call a "clean" representation! We need only eliminate u and q. Well, that's a little bit of a lie; there can be some complications arising from the fact that we can't tell the equations we want there to be _real_ solutions u and q (as opposed to complex ones) but let's just ignore that little detail (and so, in the process, we may pick up some "spurious points"). Elminating q is easy, leaving two polynomials in x, y, and u. Eliminate u. This is rather lengthy to compute with Maple, and certainly a mess to present. You can get a sense of what kind of curve you have by taking a sample pair of values for a, b. Here's the curve when a=5, b=7: 2 2 4 8 2 2 -99532800 y - 3981312 x + 764411904 - 3160512 x + 625 y + 7229952 y x 6 4 2 8 2 6 4 4 - 11956 x - 312204 x y + 2401 x + 7252 y x + 7926 x y 4 2 4 6 6 2 - 336252 y x + 4651776 y - 91300 y + 3700 y x = 0 (Aren't you glad you asked?) I'll save you the trouble retyping that: it's -99532800*y^2-3981312*x^2+764411904-3160512*x^4+625*y^8+7229952*y^2*x^2-11956* x^6-312204*x^4*y^2+2401*x^8+7252*y^2*x^6+7926*x^4*y^4-336252*y^4*x^2+4651776*y ^4-91300*y^6+3700*y^6*x^2=0; Note that only even powers of x and y are involved, reflecting the symmetry of the curve about the axes. The general case presents the curve similarly as (a component of) the zero locus of a single polynomial equation which is of degree 4 in x^2 and y^2. The coefficients are quite a mess. I'm not sure what else to tell you about the curves obtained this way. As I pointed out, the curve you want is just one component of this curve. I did some other quick computations on the curve shown above. There are singular points along the axes which are not on the component you're interested in. The curve has genus 1, meaning you can't parameterize this curve itself using rational functions of a single variable. You can traverse each quadrant by expressing x and y as _square roots_ of rational functions of one variable, though. (The curve obtained by setting x^2 = X and y^2 = Y has genus zero, and just one singular point, having Y = 47.77418... and X = -4.873806...) By the way, what you've created is what's known as the "envelope" of the original ellipse. These are discussed in some texts on Groebner basis techniques in algebraic geometry. The answers are almost never "pretty". dave