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Topic: 3)MIT math dept all fired if unable to teach correct math-- Conic
section = Oval, never ellipse//clamors Alouatta

Replies: 4   Last Post: Oct 2, 2017 7:58 AM

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 Me Posts: 1,716 Registered: 1/23/16
Re: 3)MIT math dept all fired if unable to teach correct math-- Conic
section = Oval, never ellipse//clamors Alouatta

Posted: Oct 1, 2017 3:32 PM

On Sunday, October 1, 2017 at 9:25:13 PM UTC+2, Me wrote:
> On Saturday, September 30, 2017 at 10:53:35 AM UTC+2, Archimedes Plutonium
> wrote: ...
>
> Let's consider the Sectioning of a Cylinder.
>
>          E
>         ___
>
>     .'       `.
>    /            \
> G .             .
> <---+------+c      | H
> y   .              .
>    \            /
>    `.         ´
>       `  ___ .´
>           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 [certain] 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. QED

>
> Actually, you DIDN'T show (prove) that the figure is an ellipse.
>
> Here's is an easy prove for it:
>
> Cylinder (side view):
>
> | | |
> |-------+-------+ <= x = h
> | | ´|
> | | ´ |
> | |´ |
> | ´ | |
> | ´ | |
> x = 0 => ´-------|-------|
> | r | |
>
> d(x) = r - (2r/h)x
>
> y^2 = r^2 - d(x)^2 = r^2 - r^2(2x/h - 1)^2 = r^2(1 - 4(x - h)^2/h^2
>
> => (1/r^2)y^2 + (4/h^2)(x - h)^2 = 1 ...equation of an ellipse
>
> Considerations:
>
> => y(h/2 + x')^2 = sqrt(r^2 - r^2(2(h/2 + x')/h - 1)^2) = r^2 - r^2(2x'/h)^2
>
> => y(h/2 + x') = r * (sqrt(1 - (2x'/h)^2) ...symmetric relative to h/2
>
> => y(h/2) = r (= Gc = cH)

Cone (side view):

.
/|\
/ | \
/b | \
/---+---´ <= x = h
/ |´ \
/ ´ | \
/ ´ | \
x = 0 => ´-------+-------\
/ a | \

r(x) = a - ((a-b)/h)x
d(x) = a - ((a+b)/h)x

y(x)^2 = r(x)^2 - d(x)^2 = ab - ab(2x/h - 1)^2 = ab(1 - 4(x - h)^2/h^2

=> (1/ab)y(x)^2 + (4/h^2)(x - h)^2 = 1 ...equation of an ellipse

Considerations:

=> y(h/2 + x')^2 = sqrt(ab - ab(2(h/2 + x')/h - 1)^2) = ab - ab(2x'/h)^2

=> y(h/2 + x') = sqrt(ab) * (sqrt(1 - (2x'/h)^2) ...symmetric relative to h/2

=> y(h/2) = sqrt(ab)

======================================================

@Archie: Yes, this proves that a cone section "as depicted in my diagram" is an ellipse.

Date Subject Author
10/1/17 Me
10/1/17 Me
10/2/17 plutonium.archimedes@gmail.com
10/2/17 Me