## Given Parabola, Find Axis

Professor McWorter brought to my attention the following problem: given a parabola, i.e., a curve in a plane which is a parabola. Use ruler and compass to find its axis.

The solution proceeds in a few steps. First, observe that only a ruler is needed to construct a tangent to a conic section at a point on that section.

Given two such tangents, they form a triangle with the base joining the points of tangency. The median to the base is shown to be parallel to the axis of the parabola. This actually a lemma due to Archimedes.

Having a line parallel to the axis, we draw a chord perpendicular to the line. The axis of the ellipse is the line perpendicular to the chord at its midpoint.

Professor McWorter credits Vitaly Bergelson of the Ohio State University with a simpler solution:

In some system of coordinates a parabola has the equation

There is a shortcut that does not require algebra. In a circle, a line through the midpoints of two parallel chords passes through the center of the circle. By projective transformation, the same holds for ellipse and, in principle, for parabola as well. However, for parabola, the center is a point at infinity together with the second focus. It thus follows, that a line through the midpoints of two parallel chords is parallel to the axis of the parabola.

### Conic Sections > Parabola

- The Parabola
- Archimedes Triangle and Squaring of Parabola
- Focal Definition of Parabola
- Focal Properties of Parabola
- Geometric Construction of Roots of Quadratic Equation
- Given Parabola, Find Axis
- Graph and Roots of Quadratic Polynomial
- Greg Markowsky's Problem for Parabola
- Parabola As Envelope of Straight Lines
- Generation of parabola via Apollonius' mesh
- Parabolic Mirror, Theory
- Parabolic Mirror, Illustration
- Three Parabola Tangents
- Three Points on a Parabola
- Two Tangents to Parabola
- Parabolic Sieve of Prime Numbers
- Parabolic Reciprocity
- Parabolas Related to the Orthic Triangle

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

65971433