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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?

Replies: 4   Last Post: Oct 3, 2017 12:10 PM

 Messages: [ Previous | Next ]
 Me Posts: 1,716 Registered: 1/23/16
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 12:10 PM

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 UTC-5, 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 - ((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)^2/h^2
> >
> > => (1/ab)y(x)^2 + (4/h^2)(x - h/2)^2 = 1 ...equation of an ellipse
> >
> > Some considerations:
> >
> > => y(h/2 + x')^2 = 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 y-direction):

(1/ab)y(x)^2 + (4/h^2)(x - h/2)^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 = 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)^2 = 1

=> (x - r)^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 - r) instead of just x in the equation above.

Date Subject Author
10/3/17 plutonium.archimedes@gmail.com
10/3/17 Me
10/3/17 plutonium.archimedes@gmail.com
10/3/17 Me
10/3/17 Me