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
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 $$x=-\frac e{1-e^2}.$$
Again, this is due to the degree of the algebraic equation.