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I am sorry if this question sounds stupid, but I have this doubt in mind ever since I was a child.

We know ellipses are symmetrical figures, having 2 axes of symmetry.

We know that an ellipse is generated from a cone when a plane cuts it at some angle say $\theta$ ( see conic section) Now the cone is not same shape at both ends (my doubt), so a perfect ellipse shouldn't be formed from cone as it would result in a distorted ellipse with only one axis of symmetry (my thinking)

I think that an ellipse that is a cylindric section is seemingly more symmetrical than the ellipse that is a conic section

How can I be satisfied that conic section ellipse is truly an ellipse? I think we need to solve the cone with a plane equation right?

Also, How can both cone and cylinder generate perfect ellipse?

enter image description here

Here in this figure, to me the ellipse looks imperfect. But really it isn't!

EDIT

Recently (August 1, 2018) a new video was added by 3B1B adressing this very question, and he used dandelin sphere proof to show equivalence of Thumbtack construction and slicing cone/ cylinder construction. Watch here: Youtube

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  • $\begingroup$ The axes of symmetry of the ellipse are not necessarily the axes of symmetry of the cone. Think about a conic section with angle almost $90$ degrees to the base. $\endgroup$
    – uniquesolution
    Apr 3, 2017 at 8:09
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    $\begingroup$ You see, beliefs and "how-can" questions are really hard to settle. $\endgroup$
    – uniquesolution
    Apr 3, 2017 at 8:12
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    $\begingroup$ That's a very old question that has some very old answers, don't you think? $\endgroup$
    – uniquesolution
    Apr 3, 2017 at 8:16
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    $\begingroup$ May be this helps :en.wikipedia.org/wiki/Dandelin_spheres $\endgroup$
    – Jaideep Khare
    Apr 3, 2017 at 8:20
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    $\begingroup$ That figure doesn't depict the ellipse of intersection: it depicts a projection onto the surface of your monitor. And the projection, incidentally, looks like a perfect ellipse to my eye. $\endgroup$
    – user14972
    Apr 3, 2017 at 10:26

1 Answer 1

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Short answer:

The intersection of a plane and a quadric (among which cones and cylindres) is a conic, described by a quadratic equation $$ax^2+by^2+2cxy+2dx+2ey+f=0.$$

The conics are 4-way symmetrical curves. This is proven by the fact that the equation can be reduced to the form

$$\lambda x^2+\mu y^2=\nu$$ by translation and rotation (except in a few degenerate cases).

In polar coordinates, a conic can be represented by the equation

$$r=\frac l{1+e\cos\theta}.$$

By evenness of the cosine function, you expect the curve to by symmetrical around the axis $y=0$. But there is another, unexpected, symmetry. Indeed, we can rewrite

$$r^2=(l-er\cos\theta)^2$$ or in Cartesian coordinates

$$x^2+y^2=l^2-2ex+e^2x^2$$ then by completing the square

$$(1-e^2)\left(x+\frac{e}{1-e^2}\right)^2+y^2=l^2+\frac{e^2}{1-e^2}.$$

So the other symmetry axis is $$x=-\frac e{1-e^2}.$$

enter image description here

Again, this is due to the degree of the algebraic equation.

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    $\begingroup$ Admittedly, this doesn't help geometric intuition a lot. $\endgroup$
    – user65203
    Apr 3, 2017 at 8:31
  • $\begingroup$ Thanks a lot for your answer. It explains a lot, but I am currently trying to find a procedure to convert a second degree expression like one you described into the reduced form as you mentioned, like in a question I asked some time back here. I also don't know from where the $\Delta$ came into play, and if you have knowledge on the method of factoring by rotation and translation, please do share! Thank you! $\endgroup$
    – jonsno
    Apr 3, 2017 at 10:46
  • $\begingroup$ @samjoe: maybe this: math.stackexchange.com/a/1769439/65203 $\endgroup$
    – user65203
    Apr 3, 2017 at 10:51
  • $\begingroup$ Thanks that does help alot. I can handle translation, but rotation is difficult for me, i can identify a standard curve, but not any general curve, but your answer helped a lot. $\endgroup$
    – jonsno
    Apr 3, 2017 at 11:53
  • $\begingroup$ There were some attempts to address the geometric intuition here: math.stackexchange.com/questions/2184505/… $\endgroup$
    – David K
    Apr 3, 2017 at 13:12

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