conic sectionsgeometry
I was trying to figure out a geometric proof for why the midpoints of parallel chords of a parabola lie on the same line which is parallel to its axis.
I searched on StackExchange and people have mentioned this property but haven't proved it. I was able to prove it algebraically but is there a proof of this property that only uses geometry and various properties of parabolas?
The key is to focus on the line DE (call it $L$), which is (provably) in the middle of the picture.
We won't even think about line BC until the very end. If we consider curves of constant distance from $L$, one passing through A, and one passing through B and C (we will prove this), the picture has a nice symmetry, bringing us so close to the solution that the final step is easy.
Proof:
Point A is at some distance from $L$. If we drop perpendiculars from A and B to $L$, we get congruent triangles (AAS) and so B is the same distance from $L$ as A. Similarly for A and C, so by transitivity B and C are the same distance from $L$.
Say the perpendiculars from B and C meet $L$ at M and N. Now define GF to be the perpendicular bisector of MN, with F on BC (but we do not yet know where or at what angle GF hits BC). Triangles MBG and NCG are congruent (SAS), so BG=CG and ∠BGM=∠CGN, giving us ∠BGF=∠CGF, which means ▵BGF and ▵CGF are congruent (SAS). This finally gives us that GF, which we already know to be perpendicular to DE, is indeed the perpendicular bisector of BC. Q.E.D.
PERHAPS THIS HELPS
I've been thinking about your problem for a while. I wouldn't say that I could solve it. However, I have an idea that I would like to share with you.
You certainly remember the Dandelin spheres which may have something to do with your conjecture. The following figure depicts two cones (a black one and a grey one). These cones share the vertical axis and a Dandelin sphere. Tangent to this sphere there is a blue plane whose edge can be seen in the figure:
The blue plane and the black cone intersect in a parabola. Also, the blue plane intersects with the other cone in a hyperbola. The parabola's focus and one of the foci of the hyperbola are common.
My conjecture is that this is how to imagine your couple of parabola and hyperbola with a common focus.
Then, I don't know how to proceed.
Best Answer
It may be difficult to find a geometric proof because, once authors have built up the machinery, a projective or analytic proof is easier.
But there is a geometric proof in Taylor's Geometry of Conics, 1873, pg 4. It's reasonably straightforward, but references Euclid.II.12, which some of us may be a bit rusty on.
There is also a good discussion on Pamfilos' Parabola Chords. It includes a proof for midpoints. For some assertions in the proof, refer to the Parabola page.
For convenience, here is a capture of Taylor's proof: