Finding a Parabola
Date: 2/5/96 at 23:43:48 From: Anonymous Subject: Calculus proof (more or less) The question given to me seems simple, but no one in my class (including BJMares (the Riverdale student who asks interesting integral questions) has been able to figure it out. Question: Find the equation of the parabola that is one unit away from X^2 at all points. So it is not X^2+1 or X^2-1 (and eventually from any other parabola). In other words, we are supposed to find the general equation for a parabola (z^2) 1 unit away from another parabola (n^2). Is there any proof that has been made for this type of problem? Is it possible to find the equation from this amount of information? By the way, it is not as easy as it seems; our Calculus teacher has been working on it for four years (and he worked with the space program and at many universities teaching calculus). Thank you for your time, Scott
Date: 6/1/96 at 22:17:56 From: Doctor Pete Subject: Re: Calculus proof (more or less) I believe that the locus of points equidistant from a given parabola is not itself a parabola. Take your case, where y=x^2 is the given parabola, and assume a parabola y=(x-h)^2+k exists such that the minimal distance between any two points on each parabola is 1. Then clearly the second parabola must be of the form ax^2 +/- 1. So suppose this is ax^2-1 (the positive case is similar). Now, we can restrict ourselves to looking at the first quadrant, since the parabolas are symmetrical. Consider (x0,x0^2), a point on the given parabola. The tangent line T0 at this point is y = 2*x0*x-x0^2. Then the point on the second parabola which is closest to (x0,x0^2) is the point (x1,a*x1^2-1) that has a tangent line T1 such that T1 is parallel to T0, if indeed the two parabolas are equidistant at all points. Then the distance between T1 and T0 is measured along the line through (x0,x0^2), (x1,a*x1^2-1) perpendicular to T1 and T0. This line has slope -1/(2*x0), and so x1 = x0+2*x0/sqrt(4*x0^2+1) a*x1^2-1 = x0^2-1/sqrt(4*x0^2+1). So if we substitute one equation into the other, we find there is no single value of a for which equality holds for all x0. It follows that no such parabola exists. Of course, this leaves the question of what exactly this locus is. But from our previous discussion, this is not too hard to determine. For each point (x0,x0^2), there is a one-to-one mapping to a point on the curve with the parameterization f(t) = (t+2*t/sqrt(4*t^2+1),t^2-1/sqrt(4*t^2+1)) which satisfies our locus in the first quadrant, and g(t) = (t-2*t/sqrt(4*t^2+1),t^2-1/sqrt(4*t^2+1)) which satisfies our locus in the second quadrant. Given this, I do not know whether or not the curve defined by the union of f(t) and g(t) can be expressed as some simple function h(x), but I suspect that it is not possible. -Doctor Pete, The Math Forum
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.