# Visualisation of ellipse as conic section

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?

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

• 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. – uniquesolution Apr 3 '17 at 8:09
• You see, beliefs and "how-can" questions are really hard to settle. – uniquesolution Apr 3 '17 at 8:12
• That's a very old question that has some very old answers, don't you think? – uniquesolution Apr 3 '17 at 8:16
• May be this helps :en.wikipedia.org/wiki/Dandelin_spheres – Jaideep Khare Apr 3 '17 at 8:20
• 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. – user14972 Apr 3 '17 at 10:26

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 $$x=-\frac e{1-e^2}.$$

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

• Admittedly, this doesn't help geometric intuition a lot. – Yves Daoust Apr 3 '17 at 8:31
• 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! – jonsno Apr 3 '17 at 10:46
• @samjoe: maybe this: math.stackexchange.com/a/1769439/65203 – Yves Daoust Apr 3 '17 at 10:51
• 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. – jonsno Apr 3 '17 at 11:53
• There were some attempts to address the geometric intuition here: math.stackexchange.com/questions/2184505/… – David K Apr 3 '17 at 13:12