The Closest Point on a Parabola to a Point on its Axis

Published by MrHonner on

closest point to x^2While playing around with distances from points to graphs, I discovered the following interesting property of parabolas.

Suppose you have a standard parabola, y = x^2, and a point on its axis of symmetry, (0,a).  Then the point on the graph of the parabola closest to the point (0,a) either has y-coordinate equal to a - \frac{1}{2}, or is the vertex of the parabola.

A calculus-based proof is fairly straightforward.

The distance between a point on the parabola, say (x, x^2) and the point (0,a) is given by

d(x) = \sqrt{x^2 + (x^2-a)^2}

To find the minimum distance, differentiate d(x) to get

d'(x) = \frac{2x + 2(x^2-a)2x}{2\sqrt{x^2 + (x^2-a)^2}}

This derivative is undefined when the denominator is zero, which only happens when the point (0,a) is the vertex.  The derivative is zero when

2x + 2(x^2-a)2x = 2x ( 1 + 2(x^2 - a)) = 2x(1 + 2x^2-2a)=0

By the zero-product property, either x = 0 or 1 + 2x^2-2a = 0.  

The critical value x = 0 corresponds to the vertex of the parabola.  The distance from (0,a) to the vertex is a minimum if a \leq \frac{1}{2}, or a relative maximum when a > \frac{1}{2}

Now, if 1 + 2x^2 - 2a = 0, we have x^2 = a - \frac{1}{2}.  Since y = x^2 on the parabola, the point associated with this critical value has y-coordiante a - \frac{1}{2}.  Some simple analysis shows that the distance to this point is minimal.

As an alternative to optimization, one could approach the problem geometrically.  Since the shortest path from a point to a line is the perpendicular line segment connecting them, the shortest path between a point and a curve should be a line segment perpendicular to a tangent to the curve.  At a point on the parabola (x,x^2), the slope of the tangent line is 2x, and so the slope of the line perpendicular to the tangent at (x,x^2) would be -\frac{1}{2x}.  This is equivalent to -\frac{\frac{1}{2}}{x}, and since x is the distance to the axis of symmetry, it follows that this line intersects the axis of symmetry at a point a half-unit up in the y-direction.

This fun and surprising result would make a nice exploration in a Calculus class.  It easily extends to general parabolas, and the situation naturally suggests some compelling extension questions.  

You can explore the problem, and create variations, in this interactive Desmos demonstration I put together.


Leave a Reply

Your email address will not be published. Required fields are marked *


Get every new post delivered to your Inbox

Join other followers: