The Reflection Property of a Hyperbola

[Evgeny.Makarov]

Hello,

Does anyone have a reference to a proof of the reflective property of a hyperbola? I need a proof that uses the geometric definition of a hyperbola as the locus of points $X$ such that $|XF_1-XF_2|=2a$ for some fixed points $F_1$ and $F_2$ and a positive constant $a$. The proof may also use elementary geometry but, preferably, no heavy algebra. The reflective property of a hyperbola says that a ray issued from one of the foci and reflected from the hyperbola is seen as issued from the other focus. I have a corresponding proof for an ellipse, but I looked through two of my textbooks and the first page of Google results and did not find a suitable proof for a hyperbola.

Thank you.

 

[ThePerfectHacker]

Classical geometry of conic section is a dead subject sadly. When I taught the conic sections before I researched classical geometric proofs of the reflection properties of all three conics and how the locus definition follows from planar intersections of a double-cone. None of which uses coordinates of algebra (which is what makes it classical geometry). It is a very beautiful subject nobody knows anything about, it died back in Greece. The book I used was "Geometry of Conics" by Akopyan to research a little bit of what I was interested in.
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We want to prove that $\angle F_1 P F_2$ is bisected by the hyperbola. Bisected by the hyperbola means that the tangent line bisects the angle (tangent line not drawn in picture). So we do it by contradiction. We assume that the hyperbola does not bisect the angle i.e. the tangent line does not bisect it. Therefore, the bisection of $\angle F_1 P F_2$ is a secant line to the hyperbola. It will intersect the hyperbola at some other point $Q$, we assume that $Q$ is on the same branch of the hyperbola. We will then need to redo the proof assuming $Q$ is on the other branch of the hyperbola, which we will not do as it will be very similar.

So here is the set up, $\angle F_1 P F_2$, the green angle is bisected by the brown line at another point $Q$ as we are assuming, for the sake of contradiction, the tangent line at $P$ does not bisect angle. Clearly, $F_1$ and $F_2$ are foci of hyperbola.

1) Construct $F$ on segment $F_1P$ so that $PF = PF_2$.
2) By SAS it follows $FQ = F_2Q$.
3) By hyperbola property, $F_1P - PF_2 = F_1Q - F_2Q$.
4) Write $PF_1 = PF + FF_1$.
5) By (2) in (3) we have $F_1P - PF_2 = F_1Q - FQ$
6) Substitute (1) and (4) into (5), $PF + FF_1 - PF = F_1Q - FQ \implies FF_1 = F_1Q-FQ$
7) By $\triangle FF_1Q$ by triangle inequality satisfies $FF_1 > F_1Q - FQ$. Contradiction!

 

[Evgeny.Makarov]

Thanks a lot.