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Topic: 3)Stanford's entire math dept. fired//unable to confirm that oval is
conic not ellipse//as insane in math as Jan Bielawski

Replies: 12   Last Post: Oct 3, 2017 6:16 AM

 Messages: [ Previous | Next ]
 plutonium.archimedes@gmail.com Posts: 18,572 Registered: 3/31/08
Re: Stanford math dept as dumb as Mike Moroney in math??// Conic
section is OVAL, never ellipse

Posted: Oct 3, 2017 12:23 AM

Franz, Me wrote:

On Monday, October 2, 2017 at 11:51:08 AM UTC-5, Dan Christensen wrote:

Mr Sawat Layuheem
11:01 PM (17 minutes ago)

Translate message to English

Triangle Absurb???????????????

Dan is ellipse in beer bottle

-------------

level of math of these two

On Monday, October 2, 2017 at 10:36:22 AM UTC-5, Me wrote:
> On Monday, October 2, 2017 at 5:08:21 PM UTC+2, Archimedes Plutonium wrote:
>

> > Analytic Geometry Proof, Cylinder Section [as well as cone section] is a[n]
> > Ellipse:

>
> ^ x
> E|
> -+- <= x = h
> .' | `.
> / | \
> . | .
> 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.
> >
> > So, what is the proof that [cut] figure EGFH is always an ellipse in the
> > cylinder section [as well as in the cone section]?

>
> It turns out that a cylinder can be considered as a special case of a cone here. Actually, there's a simple proof which works for both cases, cone and cylinder.
>
> (@Archie: This shows that there is no essential difference between these two cases.)
>
> Cone/Cylinder (side view):
>
> / | \
> /b | \
> /---+---´ <= x = h
> / |´ \
> / ´ | \
> / ´ | \
> x = 0 => ´-------+-------\
> / a | \
>
> (cone: b < a; cylinder: a = b = r)
>
> 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
>
> Some 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 (hence Ec = cF)
>
> => y(h/2) = sqrt(ab) (= Gc = cH)
>
> ======================================================
>
> @Archie: Yes, this proves that (certain) cone sections "as depicted in my diagram" as well as (certain) cylinder sections (as described by you) are ellipses. qed
>
> Note, Archie, that there is no reference to Dandelin Spheres whatsoever.
>
> Still not convinced? Can you point out an error in my simple calculation (of the shape of the cone/cylinder section) above?