# The Closest Point on a Parabola to a Point on its Axis While 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.

Categories: GeometryTeaching