Notes on tangents to parabolas

Transcription

1 Notes on tangents to parabolas (These are notes for a talk I gave on 2007 March 30.) The point of this talk is not to publicize new results. The most recent material in it is the concept of Bézier curves, which ates to the late 1950s; the secon most recent material is the quarature of the parabola, which is a famous result of Archimees from 250 BC or so. The point of this talk is, rather, to avertise synthetic geometry, that is, geometry without coorinates, as practice by Eucli. Consier, for example, the following proposition: Let the tangents to a parabola at an meet at. Let M an N be the mipoints of an respectively. Then MN is also tangent to the parabola, an the point of tangency is the mipoint of MN. M N We ll prove this below. For now, just notice how thoroughly routine it is to prove it with moern techniques. Aopting an appropriate coorinate system, the parabola has equation y = x 2. Let be the point (p, p 2 ), an let be the point (q, q 2 ). Now just compute everything: eploy the techniques of firstyear calculus to fin the equations of the tangent lines; solve a two-equation linear system to fin the coorinates of their intersection; an so on. Those computations are not totally without interest, but they on t leave me feeling that I unerstan the phenomenon. A synthetic approach yiels, at least, an instructively ifferent view of it. The first orer of business is to unerstan what a tangent is, a question that turns out to be somewhat thornier than you might expect. In calculus we have a efinition of tangents (of graphs of functions, at first, an later of more general curves) in terms of erivatives, but this requires a large apparatus of theory which involves coorinates at a funamental level, an is rather in the wrong spirit for what we re oing. Steven Taschuk 2011 August

2 In Eucliean geometry one usually eals with tangents to circles, an for circles we have a simple efinition: a line is tangent to a circle if it meets the circle at exactly one point. This efinition will not, however, serve for parabolas. We o inee expect that a tangent to a parabola will meet the parabola only at the point of tangency, but so too will a line parallel to the axis of the parabola. 1 Eucli s own efinition of tangents is slightly more helpful: A straight line is sai to touch a circle which, meeting the circle an being prouce, oes not cut the circle. 2 That is, a tangent is a line that meets but oes not cut. Meeting is easy to unerstan: lines an other curves are sets of points, an they meet at a point if they both contain that point. Cutting is a bit more mysterious; Eucli oesn t efine it. The simplest interpretation I can think of is this: a line ivies the plane into two regions; the line, we say, cuts a figure if the figure contains points in both of those regions. 3 There are a few ways to efine parabolas; for our purposes, the simplest one is this: We are given a point F, calle the focus of the parabola, an a line, calle its irectrix. The parabola consists of those points which are equiistant from the focus an the irectrix. 1 The righteous way to eal with this problem is actually foun in projective geometry; there, the parabola an a line parallel to its axis o meet at a secon point, a point at infinity, so we can efine tangents for parabolas as we o for circles. (Inee, in projective geometry all conics are the same.) 2 Eucli s Elements, Book III, Definition 3, quote from The Thirteen Books of Eucli s Elements, trans. an e. Thomas L. Heath, 2n e. (New York: Dover, 1956), 2:1. 3 This efinition of cut means that a line that meets but oes not cut is, in the terminology of convex geometry, a supporting hyperplane, not a tangent. We can get away with this because the interior of the parabola is convex. (roving that is a goo exercise.) Steven Taschuk 2011 August

3 F 0 As usual, when we speak of istance between a point an a line, we mean the perpenicular istance. Thus, to etermine whether a point is on the parabola, we rop the perpenicular from to, meeting at 0 ; then we compare the istance F to the istance 0. is on the parabola if an only if F = 0. We will be ropping perpeniculars to quite often in what follows. The foot of the perpenicular will always be name with the prime symbol; that is, for any point X, the foot of the perpenicular roppe from X to will be calle X 0. Now, the fact that a point on the parabola is equiistant from 0 an F has a familiar consequence: lies on the perpenicular bisector of the segment 0 F. Our first theorem asserts that this line not only passes through but is the tangent to the parabola there. One-Tangent Theorem Let lie on the parabola. The perpenicular bisector of 0 F is the tangent to the parabola at. roof Let l be the perpenicular bisector of 0 F. We have alreay seen that l passes through, so l meets the parabola. It remains to show that l oes not cut the parabola. We will show that every point on the parabola is on the same sie of l as F; equivalently, that no point on the opposite sie of l from F is on the parabola. Steven Taschuk 2011 August

4 l F 0 0 Let lie on the opposite sie of l from F. Join F, meeting l at. We have F = + F ( is between an F) = + 0 ( is on l) > 0 (triangle inequality; see below) 0 (perpenicular is shortest) an so oes not lie on the parabola. The inequality + 0 > 0 is strict because oes not lie on the segment 0. For since l is, by construction, the perpenicular bisector of 0 F, 0 an F lie on opposite sies of l. Therefore an 0 lie on the same sie of l, that is, the segment 0 oes not intersect l. In particular, the segment 0 oes not contain. This proof is, in fact, not complete. It oes show that, by our efinitions, the line escribe is tangent to the parabola at ; but the theorem asserts that it is the tangent to the parabola at. We have not seen any argument that the parabola has only one tangent at each point. The only proof I have of this statement (in the style of geometry we re oing) is quite long an involve, so I omit it. If you know of a simple proof, I like to hear about it. As an application of the One-Tangent Theoα rem, we can prove the famous optical property of the parabola. In the figure at right, α = β since γ F the tangent bisects \F 0, an β = γ as vertical angles. So α = γ, which since light bounces β with the angle of incience equal to the angle of reflection entails that a beam of light emanating from F an hitting the parabola at will bounce 0 off along the (extension of the) line 0. An, of course, in reverse: light arriving perpenicular to the irectrix will be reflecte to the focus. Steven Taschuk 2011 August

5 The following theorem gives another example of how the One-Tangent Theorem turns knowlege of perpenicular bisectors into knowlege of tangents to parabolas. Two-Tangent Theorem Let an lie on the parabola, an let the tangents at an meet at. Then 0 bisects 0 0. roof By the One-Tangent Theorem, is the perpenicular bisector of 0 F, an is the perpenicular bisector of 0 F. Therefore their intersection is the circumcentre of 4 0 F 0, an 0, being the perpenicular from to 0 0, is also the perpenicular bisector of 0 0. F Three-Tangent Theorem Let,, an X lie on the parabola. Let the tangents at an meet at. Let the tangent at X meet at M an at N. Then roof M M = MX XN = N N. M X N 0 M 0 X 0 0 N 0 0 Steven Taschuk 2011 August

6 By the Two-Tangent Theorem (for the tangents M an MX), 0 M 0 = M 0 X 0. Furthermore, 0 M 0 = X 0 (Two-Tangent Th.: M, MX) = X 0 0 = 0 0 N 0 0 (Two-Tangent Th.:, ; an XN, N) = 0 N 0. Thus 0 M 0 = M 0 X 0 = 0 N 0. Similarly, M 0 0 = X 0 N 0 = N 0 0, so 0 M 0 M 0 0 = M 0 X 0 X 0 N 0 = 0 N 0 N 0 0. The esire equalities now follow by parallels. Letting t = M (so that the proportion mentione in the statement of the Three-Tangent Theorem is (1 + t) 1 ) an consiering the points as vectors (so we can apply vector arithmetic), we have whence M = (1 t) + t, N = (1 t) + t, an X = (1 t)m + tn, X = (1 t) 2 + 2t(1 t) + t 2. As t varies from 0 to 1, the point X varies along the parabola from to. Notice that the coefficients of this linear combination of,, an arise from the expansion of the square of a binomial: ((1 t) + t) 2 = (1 t) 2 + 2t(1 t) + t 2. It is, then, natural to consier analogous parameterizations of curves base on the expansion of other powers of a binomial. For example, consiering the cube of a binomial leas to the curve (1 t) 3 u + 3t(1 t) 2 v + 3t 2 (1 t)w + t 3 x (where u, v, w, an x are fixe vectors). The families of curves obtaine by such parameterizations are known as Bézier curves, an are much use in computer graphics an computer-aie esign. 4 The Three-Tangent Theorem can also be applie to prove a famous result of Archimees. 5 Taking the special case where the proportion in the theorem 4 A great eal of information about these curves can be foun by web search. 5 See his uarature of the arabola, in The Works of Archimees, trans. an e. Thomas L. Heath (New York: Dover, 2002), He gives two proofs; the one in the text iffers from both. Also, his statement of the result is that the area of the parabolic segment is 4 the area of 4X. 3 Steven Taschuk 2011 August

7 is 1 yiels the corollary mentione at the beginning of these notes: Let the tangents to a parabola at an meet at. Let M an N be the mipoints of an respectively. Then MN is also tangent to the parabola, an the point of tangency X is the mipoint of MN. M X N Let the area of 4, the triangle forme by the two tangents an the secant, be 1. The parabolic segment X inclues the triangle 4X, with area 1 2, an exclues the triangle 4MN, with area 1 4. The remaining pieces, 4MX an 4XN, are two smaller instances of the same situation: they are triangles forme by two tangents an a secant. By the analogous subivision of these triangles, the parabolic segment inclues 1 2 of their area an exclues 1 4 of it, leaving four new smaller instances of the same situation again. Continuing in this manner, we see that the area of the parabolic segment is 1 2 /( ) = 2 3 of the area of 4. Steven Taschuk 2011 August