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Topic:
(5)are all Univ.Western Ontario mathematicians like Ajneet Dhillon, Matthias Franz, John Jardine as dumb as Dan Christensen, teaching a Conic section is ellipse, when in t ruth it is an oval?
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Last Post:
Oct 3, 2017 12:10 PM



Me
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Re: (5)are all Univ.Western Ontario mathematicians l ike Ajneet Dhillon, Matthias Franz, John Jardine as dumb as Dan Christensen, teaching a Conic section is ellipse, when in truth it is an oval?
Posted:
Oct 3, 2017 10:15 AM


On Tuesday, October 3, 2017 at 2:45:54 PM UTC+2, Archimedes Plutonium wrote: > On Tuesday, October 3, 2017 at 5:04:59 AM UTC5, Me wrote: > > > > 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/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) > > > > ====================================================== > > > Franz, do you see that a oval is distinct from a ellipse
Not necessarilly, since the ellipse is (a special case of) an oval. But I get your point. Following your definition of an oval the ellipse is not an oval. Right.
Still, proof / calculation above gives the following EQUATION for the cut (in ydirection):
(1/ab)y(x)^2 + (4/h^2)(x  h)^2 = 1 (*)
This is a equation of an ellipse. Actually, we can PROVE that there's a symmetrie relative to x = h/2:
=> 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 x = h/2
=> Hence Ec = cF (!)
^ x E + <= x=h .'  `. /  \ .  . G  +c  H .  . \  / `.  .´ y <+ <= x=0 F
With other words, GH is an axis of symmetrie of the figure (besides EF).
We can then compute Gc and cH:
=> y(h/2) = sqrt(ab) (= Gc = cH)
Now IF for a cone or cylinder b = a = r and h = 2r (i.e. the cutting plain is perpenticular to the axis of the cone/cylinder), then the cut figure (cone/cylinder section) is a circle.
We can get this immediately from our equation (*):
=> (1/r^2)y^2 + (1/r^2)(x  h)^2 = 1
=> (x  2r)^2 + y^2 = r^2
Note that the origin of our coordinate system doesn't lie in the center of the circle, hence we have (x  2r) instead of just x in the equation above.



