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Topic:
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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11
Last Post:
Oct 4, 2017 7:05 PM




Princeton math dept agreeing with Franz for none has spotted the error as dumb as Franz, teaching a Conic section is ellipse, when in truth it is an oval?
Posted:
Oct 2, 2017 11:00 AM


On Monday, October 2, 2017 at 1:37:38 AM UTC5, Me wrote: > Let's consider the Sectioning of a Cylinder and a Cone. > > ^ x > 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. > > > > So, what is the proof that [cut] figure EGFH is always an ellipse in the > > cylinder section [as well as in the cone section]? > > Here's is an easy proof 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  ((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/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) (= Gc = cH) > > ====================================================== > > It turns out that a cylinder can be considered as a special case of a cone here. Actually, the latter proof works for both cases, cone and cylinder. > > Cone/Cylinder (side view): > > /  \ > /b  \ > /+´ <= x = h > / ´ \ > / ´  \ > / ´  \ > x = 0 => ´+\ > / a  \ > > (cone: b < a; cylinder: a = b = 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/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) (= 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?
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