# Pure geometry proof of parabolic mirror property

Is there a proof of the property that parallel rays of light incident upon a parabolic mirror converge to its focus that does not resort to Cartesian coordinates?

• There is. But it requires a shift in your thinking away from what physicists are comfortable with. We think of a parabola as the curve defined by functions like y=x^2. You want to think in terms of geometry, where a parabola is the intersection of a plane and a cone where the axis of the cone is parallel to the plane. This question might do better on the math site. Commented Aug 12, 2017 at 4:56
• You might also read "The Archimedes Codex" It goes through some of the math used by Archimedes. One of the problems involves a parabola, where we would use Cartesian coordinates, and of course he did not. It's a good read. Commented Aug 12, 2017 at 5:00
• Have a look here and remember that the parallel rays travel equal distances to reach the focus and so arrive in phase with one another. jwilson.coe.uga.edu/EMAT6680Fa08/Wisdom/EMAT6690/Parabolanjw/… Commented Aug 12, 2017 at 5:35
• Conic Sections ? Math SE. Commented Aug 12, 2017 at 10:18
• Here is an attempt with the formula for the equation of the line reflected by another line + using the equation of the tangent of the parabola x^2 with its derivative: math.stackexchange.com/questions/4852497/…
– Basj
Commented Jan 28 at 7:15

Physicists normally think of parabolas as the locus of equations of the form $y=x^2$, but you're explicitly looking for something with no formulas, so you probably need to change that as well. The next likeliest candidate is defining the parabola as the locus of points equidistant from a point and a line, which ties you in to all the proofs of classical geometry. Under that understanding, the proof of the reflection property is a staple of euclidean compass-and-straightedge geometry, and Wikipedia has a suitable proof, based on a construction of the form
In short, with $C$ on the directrix, you define $B$ as the midpoint of $\overline{FC}$, which means (since $\overline{FE}=\overline{EC}$ by definition of the parabola) that $\angle FEB=\angle BEC$, so you just need to show that the line joining $B$ and $E$ is tangent to the parabola. The Wikipedia proof relies on some facts from calculus, though if you want a calculus-free proof you can probably find one by rooting around in the toolbox for the euclidean geometry of conic sections.