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6)are all German mathematicians like Peter Roquette, Gunther Schmidt KarlOtto Stöhr as dumb as Franz, teachi ng a Conic section is ellipse, when in truth it is an oval?
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Oct 4, 2017 7:05 PM



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Re: 6)are all German mathematicians like Peter Roque tte, Gunther Schmidt KarlOtto Stöhr as dumb as Franz, te aching a Conic section is ellipse, when in truth it is an ov al?
Posted:
Oct 4, 2017 7:05 PM


On Wednesday, October 4, 2017 at 10:29:48 PM UTC+2, Jan wrote:
> Secondly, ellipse is a conic section. Here is an elegant proof: > https://en.wikipedia.org/wiki/Dandelin_spheres
Here is another on, less elegant but still helpfull, I'd say, since it proves explicitely some of your claims below.
Top view of the cone section and depicion of the coordinate system used in the proof below:
^ x  + <= x=h .'  `. /  \ .  .    .  . \  / `.  .´ y <+ <= x=0
Cone/Cylinder (side view): /  \ (with b <= a) /b  \ /+´ <= x = h / ´ \ / ´  \ / ´  \ x = 0 => ´+\ / a  \
(cone: b < a, cylinder: b = a = r)
r(x) = a  ((ab)/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)^2/h^2
=> (1/ab)y(x)^2 + (4/h^2)(x  h/2)^2 = 1 ...equation of an ellipse
qed.
Now lets just look at some "properties" of this "curve" (i.e. the cone/cyliner section):
y(h/2 + x')^2 = ab  ab(2(h/2 + x')/h  1)^2 = ab  ab(2x'/h)^2
=> y_1(h/2 + x') = sqrt(ab) * (sqrt(1  (2x'/h)^2) ...symmetric relative to x = h/2 (y_1(x) ist the part of the curve with y >= 0)
=> y_2(h/2 + x') = sqrt(ab) * (sqrt(1  (2x'/h)^2) ...symmetric relative to x = h/2 (y_2(x) ist the part of the curve with y <= 0)
And clearly there's a "maximum" of y_1(x) and a "minimum" of y_2(x) at x = h/2.
With other words, x = h/2 is another axis of symmetry of the figure besides y = 0.
So it's clearly NOT an "oval" as defined by AP.
> ... it [the cone section] has two axes of symmetry. There are more subtle > reasons for that (to actually see it, you need an actual proof [...] > but one thing you overlooked is that the entire crosssectional area > shifts sideways as you rotate the sectional plane. > > You assume the cone's centreline must intersect the crosssection in > its centre of symmetry but this is false. (It happens top be true in > the case of the cylinder which is probably what led you into this trap.) > > [...] All you have done is you've shown that a particular method does > not detect more than one axis of symmetry. In order to PROVE your claim > you have to PROVE there is no OTHER axis of symmetry. Your proof fails > to do that, it simply detects one and it does not DISPROVE the existence > of any other. > > In reality, of course, the crosssection figure does have a second axis > of symmetry. Your proof is simply too weak to detect it.



