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Topic: Array:: Analytic Geometry proof that Cylinder section= Ellipse//Conic
section = Oval, never ellipse

Replies: 1   Last Post: Sep 30, 2017 3:53 AM

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 plutonium.archimedes@gmail.com Posts: 18,572 Registered: 3/31/08
Array:: Analytic Geometry proof that Cylinder section= Ellipse//Conic
section = Oval, never ellipse

Posted: Sep 30, 2017 3:06 AM

Alright, I gave a proof, a Synthetic Geometry proof that the Conic Section was a Oval, never an ellipse and the Cylinder section turns out to be the ellipse.

A Synthetic geometry proof is where you need no numbers, no coordinate points, now arithmetic, but just using concepts and axioms. A Analytic Geometry proof is where numbers come into the picture.

So, here is a Analytic Geometry proof that the Cylinder Section is a Ellipse, and a Conic Section is a Oval, never an ellipse.

Now I did 3 Experiments and 3 models of the problem, but it turns out that one model is superior over all the other models. One model is the best of all.

That model is where you construct a cone and a cylinder and then implant a piston circle inside the cone and cylinder attached to a handle so that you can rotate the circle inside. Mine uses a long nail that I poked holes into the side of a cylinder and another one inside a cone made from heavy wax paper of magazine covers. And I used a Mason or Kerr used lid and I attached them to the nail by drilling two holes into each lid and running a wire as fastener. All of this done so I can rotate or pivot the circle inside the cylinder and con. You need a long nail, for if you make the models too small or too skinny, you lose clarity.

ARRAY, Analytic Geometry Proof, Cylinder Section is a Ellipse::

E
__
.-'              `-.
.'                    `.
/                         \
;                           ;
| G          c             | H
;                           ;
\                         /
`.                     .'
`-.    _____  .-'
F

The above is a view of a ellipse with center c and is produced by the Sectioning of a Cylinder as long as the cut is not perpendicular to the base, and as long as the cut involves two points not larger than the height of the cylinder walls. What we want to prove is that the cut is always a ellipse, which is a plane figure of two axes of symmetry with a Major Axis and Minor Axis and center at c.

Side view of Cylinder EGFH above with entry point cut at E and exit point cut at F and where c denotes the central axis of the cylinder and where x denotes a circle at c parallel with the base-circle of cylinder

| |
| | E
| |
| |
|x c |x
| |
| |
| |
|F |
| |
| |
| |

So, what is the proof that figure EGFH is always an ellipse in the cylinder section? The line segment GH is the diameter of the circle base of cylinder and the cylinder axis cuts this diameter in half such that Gc = cH. Now we only need to show that Fc = cE. This is done from the right triangles cxF and cxE, for we note that by Angle-Side-Angle these two right triangles are congruent and hence Fc = cE, our second axis of symmetry and thus figure EGFH is always an ellipse.

AP