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Topic: Synthetic Geometry proof and Analytical Geometry proof that Conic is
a Oval, never an ellipse// yes Apollonius was wrong

Replies: 5   Last Post: Oct 7, 2017 1:48 AM

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Posts: 1,409
Registered: 1/23/16
Re: Synthetic Geometry proof and Analytical Geometry proof that Conic
is a Oval, never an ellipse// yes Apollonius was wrong

Posted: Oct 2, 2017 2:11 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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)
- hide quoted text -

Cone (side view):

.
/|\
/ | \
/b | \
/---+---´ <= x = h
/ |´ \
/ ´ | \
/ ´ | \
x = 0 => ´-------+-------\
/ a | \

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

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)

======================================================

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 - ((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

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 (certain) cone sections "as depicted in my diagram" as well as (certain) cylinder sections (as described by you) are ellipses. qed

=> Appolonius was right.

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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