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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.


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,

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

Associated Topics:
High School Conic Sections/Circles
High School Geometry

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