Some years ago a mathematician challenged a friend of mine to prove that a parabola is symmetric without using any coordinates. My friend got stumped, and challenged me to do this problem. I also got stumped.
I tried a couple of different ways. The first was to draw a picture of a parabola and simply notice that it was symmetric, but I think knowing how to draw the parabola in the first place might have assumed coordinates. The second approach I took was to parametrize by arc length and show reflective symmetry by a reflection in arc length, but then assuming (1) that there was a coordinate systems to parametrize the curve from and (2) that putting things in terms of arc length isn't its own kind of coordinates.
I am not sure I understand what a parabola is without some understanding of coordinates. The equation $y = ax^2$ to me entails a subset of a Cartesian product $Y \times A \times X$, or a permutation thereof, which is a space that the parabola sits within.
How do I show that a parabola is symmetric without the existence of any coordinates?
You'll need to refer to lengths, but not necessarily coordinates. By definition a parabola is the curve defined by all points equidistant from a single focal point and a line, the directrix. Let's call the focus A and the line L.
Drop a line from A to L that is perpendicular to L. Call this line M and the point of intersection Q. Suppose also we have a point P on a parabola on one side of M.
With Compass centered at A, construct a circle through P. With Compass centered at Q, also construct a circle through P. Call their point of intersection R.
Now construct line segments A to P, and P to Q. Combined with L, this creates triangle APQ. Then construct line segments A to R and R to Q. This is triangle ARQ. AQ is congruent to itself. AP is congruent to AR since both are a radius of the same circle. RQ and PQ are likewise congruent. These triangles are congruent by SSS.
Now construct a line between P and R and call the point of intersection with M point T. Now we have triangles ATP and ATR. AT is congruent to itself. Angle RAT is congruent to and PAT by the previous SSS argument and the definition of congruent triangles. AR and AP are also congruent. So ATP and ATR are congruent triangles by SAS. This implies TP is congruent to TR since corresponding sides of congruent triangles are congruent. We also have that angles ATP and ATR are congruent. This implies they are right angles.
QRT and QPT are then right triangles. Further, they are congruent by HL. From this it follows that R and P are the same height above the directrix and equal and opposite distances about line M. Thus we have established the expected symmetry.
This reasoning applies for any point P on the parabola and so applies to the parabola in general.