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Topic:
3)MIT math dept all fired if unable to teach correct math Conic section = Oval, never ellipse//clamors Alouatta
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Oct 2, 2017 7:58 AM



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



