three theorems on parabolas

In the Cartesian plane, pick a point with coordinates (0,2f) (subtle hint!) and construct (1) the set S of segments s joining F=(0,2f) with the points (x,0), and (2) the set B of right-bisectors b of the segments sS.

Theorem 1 :

The envelope described by the lines of the set B is a parabolaPlanetmathPlanetmath with x-axis as directrix and focal length |f|.


We’re lucky in that we don’t need a fancy definition of envelope; considering a line to be a set of points it’s just the boundary of the set C=bBb. Strategy: fix an x coordinate and find the max/minimum of possible y’s in C with that x. But first we’ll pick an s from S by picking a point p=(w,0) on the x axis. The midpointMathworldPlanetmathPlanetmathPlanetmath of the segment sS through p is M=(w2,f). Also, the slope of this s is -2fw. The corresponding right-bisector will also pass through (w2,f) and will have slope w2f. Its equation is therefore




By any of many very famous theoremsMathworldPlanetmath (Euclid book II theorem twenty-something, Cauchy-Schwarz-Bunyakovski (overkill), differential calculus, what you will) for fixed x, y is an extremumMathworldPlanetmath for w=x only, and therefore the envelope has equation


I could say I’m done right now because we “know” that this is a parabola, with focal length f and x-axis as directrix. I don’t want to, though. The most popular definition of parabola I know of is “set of points equidistant from some line d and some point f.” The line responsible for the point on the envelope with given ordinate x was found to bisect the segment sS through H=(x,0). So pick an extra point QbB where b is the perpendicular bisectorMathworldPlanetmath of s. We then have FMQ=QMH because they’re both right anglesMathworldPlanetmathPlanetmath, lengths FM=MH, and QM is common to both trianglesMathworldPlanetmath FMQ and HMQ. Therefore two sides and the angles they contain are respectively equal in the triangles FMQ and HMQ, and so respective angles and respective sides are all equal. In particular, FQ=QH. Also, since Q and H have the same x coordinate, the line QH is the perpendicularPlanetmathPlanetmathPlanetmath to the x-axis, and so Q, a general point on the envelope, is equidistant from F and the x-axis. Therefore etc.


Because of this construction, it is clear that the lines of B are all tangentPlanetmathPlanetmathPlanetmath to the parabola in question.

We’re not done yet. Pick a random point P outside C (“inside” the parabola), and call the parabola π (just to be nasty). Here’s a nice quicky:

Theorem 2 The Reflector Law:

For Rπ, the length of the path PRF is minimalPlanetmathPlanetmath when PR produced is perpendicular to the x-axis.


Quite simply, assume PR produced is not necessarily perpendicular to the x-axis. Because π is a parabola, the segment from R perpendicular to the x-axis has the same length as RF. So let this perpendicular hit the x-axis at H. We then have that the length of PRH equals that of PRF. But PRH (and hence PRF) is minimal when it’s a straight line; that is, when PR produced is perpendicular to the x-axis.


Hey! I called that theorem the “reflector law”. Perhaps it didn’t look like one. (It is in the Lagrangian formulation), but it’s fairly easy to show (it’s a similarMathworldPlanetmathPlanetmath argumentMathworldPlanetmath) that the shortest path from a point to a line to a point makes and “reflected” angles equal.

One last marvelous tidbit. This will take more time, though. Let b be tangent to π at R, and let n be perpendicular to b at R. We will call n the to π at R. Let n meet the x-axis at G.

Theorem 3 :

The radius of the “best-fit circle” to π at R is twice the length RG.


(Note: the ’s need to be phrased in terms of upper and lower bounds, so I can use the sandwich theorem, but the proof schema is exactly what is required).

Take two points R,R on π some small distanceMathworldPlanetmath ϵ from each other (we don’t actually use ϵ, it’s just a psychological trick). Construct the tangent t and normal n at R, normal n at R. Let n,n intersect at O, and t intersect the x-axis at G. RF,RF. Erect perpendiculars g,g to the x-axis through R,R respectively. RR. Let g intersect the x-axis at H. Let P,P be points on g,g not in C. Construct RE perpendicular to RF with E in RF. We now have

  • i)


  • iii)


  • v)

    RRE+EROπ2 (That’s the number π, not the parabola)

  • vii)


  • ix)


  • xi)


  • xiii)


  • xv)


From (iii),(iv) and (i) we have RREGRH, and since R is close to R, and if we let R approach R, the approximations approach equality. Therefore, we have that triangle RRE approaches similarity with GRH. Therefore we have RR:ERRG:RH. Combining this with (ii),(vi),(vii), and (viii) it follows that RO2RG, and in the limit RR, RO=2RG.


This last theorem is a very nice way of short-cutting all the messy calculus needed to derive the Schwarzschild “Black-Hole” solution to Einstein’s equations, and that’s why I enjoy it so.

Title three theorems on parabolas
Canonical name ThreeTheoremsOnParabolas
Date of creation 2013-03-22 12:40:51
Last modified on 2013-03-22 12:40:51
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 19
Author CWoo (3771)
Entry type Topic
Classification msc 51N20
Related topic PropertiesOfParabola