# how do the conic sections add to the possibilities of geometric construction?

If we are limited by what we can construct with compass and straight edge, then what becomes possible by expanding our toolkit to include all conic sections? The tools would be based on already constructed points, lines and circles.

I'm aware that the hyperbola allows for angle trisection and the construction of a heptagon; however, I have not read anything about a conic section being used to "square the circle".

Does an ellipse or parabola contribute anything?

To be clear, my interest is not about methods for "squaring the circle" and such; it's about what the conic sections do and do not add to the realm of possibilities regarding geometric construction.

• They can only add algebraic numbers, so no circle squaring could be done. Commented Jan 25, 2015 at 3:00

## 1 Answer

You can do better than asking for conic sections in the plural. You can get the full possible extension of Euclidean constructions (meaning you can solve cubic and quartic equations) by adding just one conic section: the parabola $$y=x^2$$. We divide this into two parts: (1) how the construction works algebraically, and (2) given a parabola on an otherwise blanks sheet of paper, how to fit it to the $$y=x^2$$ paradigm.

The algebra

Suppose you have a parabola characterized as $$y=x^2$$ in the coordinate plane. Construct a circle through $$(0,0)$$ with point $$(a,b)$$ at the opposite end of the diameter through the origin. Then

$$y=x^2$$

$$x^2+y^2=ax+by$$

and upon eliminatinhg $$y$$:

$$x^4+(1-b)x^2-ax=x[x^3+(1-b)x-a]=0.$$

Thus any equaton $$x^3+px+q=0$$ is soluble if $$p$$ and $$q$$ are rational or constructible from rationals. This can be generalized to all cubics with rational or constructible coefficients through translation and then to quartics by way of their resolvent cubics. QED.

The geometry

So if you are given just the parabola and no coordonate plane, how do you render it as $$y=x^2$$? If you can constuct the directrix which would be $$y=-1/4$$ and the focus at $$(0,1/4)$$ you're there, because a parabola is uniquely defined by these features.

Consider a circle. If you draw a series of parallel chords and mark off their midpoints, you probably recognize that the midpoints fall along a diameter of the circle, called the conjugate diameter of the set of parallel chords. If this conjugate diameter is extended to intersect the circle, then the tangent to the at the point of intersection is parallel to the associated set of parallel chords. Not as well known is the fact that this property of conjugate diameters carries over to all conic sections. So given our parabola, we construct two parallel chords, connect their midpoints and, having identified the conjugate diameter, we draw the tangent parallel to the two chords through its point of intersection with the parbola.

What does that have to do with the directrux or focus? We bring in another property of a parabola: any two perpendicular tangents intersect on the directrix. So we now construct another pair of parallel chords, perpendicular to the first pair, and use their midpoints to define a second conjugate diameter and a corresponding tangent, which will indeed be perpendicular to the first tangent. There is a shortcut: to construct the midoints of the first two chords we probably used perpendicular bisectors, and now those perpendicular bisectors can be used to define the second set of chords!

We now have a pair of perpenducular tangents which intersect at one point on the directrix. But we need two points to define a line — unless we can identify a known parallel or perpendicular line. Those conjugate diameters constructed above look suspiciously parallel to the axis of the parabola. That's because they are. So given a point on the directrix from the intersection of the constructed perpendicular tangents, construct the perpendulicular from there to any known conjugate diameter. Thence, the directrix.

There are now several approaches to obtain the focus. One is as follows: with the directrix known, construct one more chord parallel to it and perpendicularly bisect it. This perpendicular bisector is, of course, the axis (and thus the $$y$$-axis of the coordinate plane), which intersects the parabola at the vertex and also intersects the directrix. From these two points the focus easily follows using the equal-distance definition of the parabola; or you can draw the line through the other two tangency points you constructed earlier and (with those tangents being perpendicular to each other) this secant intersects the axis on the focus.

With that, you're good to go with the algebra.

• Thank you. I can't believe someone finally answered this. :) Commented Dec 3, 2023 at 2:45