Nature of relationship between a cone and a pair of planes

Question

A cone is defined as a surface generated by lines passing through a fixed point and interesecting a given conic, or touching a given surface.
A pair of non-parallel planes seems to loosely satisfy these properties. We can loosely call a pair of planes as a cone with an infinite number of vertices (any point on the line of intersection of the planes being a vertex) and the guiding curve being two non-intersecting lines meeting at infinity. The pair of planes satisfies the property of cone which says that any line joining the vertex with a point of the cone has to lie on the cone. (being a generating line)
I understand that most of my claims are hand wavy but I doubt that this is just a coincidence. So the question is mathematically speaking, what is the relationship between cones and pairs of planes? Kindly explain in simple language. I am familiar with only undergraduate level mathematics.

Answer 1

Actually, cones enjoy different definitions.

1 - In linear algebra, a cone is a subset of a real vector space that is closed under multiplication by a positive scalar, cf. Wikipedia: https://en.m.wikipedia.org/wiki/Convex_cone

So yes, with this definition two non-parallel planes make a cone.

Apart from being one of Wikipedia's definitions for a cone (and they add "sometimes called a linear cone for distinguishing it from other sorts of cones"), this is also the definition used in conic optimization, cf. https://en.m.wikipedia.org/wiki/Conic_optimization

This closedness by positive multiplication is actually usefull in proofs related to optimization algorithms, i.e. it is a well-thought concept; although usually coupled with convexity: cones considered in optimization are convex cones.

2 - That being said, two non-parallel planes are also a cone according to the definition you adopt, which is also widely used, i.e. "lines passing through a fixed point and intersecting a given conic": two non-parallel planes are equal to lines passing through a fixed point and two intersecting lines, and two intersecting lines are a (degenerate) conic: https://en.m.wikipedia.org/wiki/Degenerate_conic

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EDIT
Following comments, here is a construction to relate a circular cone and two planes.
The idea is: as two intersecting lines are a degenerate conic, they can be parameterized in the same way as other conics.

We'll use as two lines in the projective plane $\mathbb P^2$ the principal axes: $x=0$ and $y=0$.
These are parameterized with $t \in [0, 2)$:
$\begin{cases} t \in [0, 1): x=\tan(\pi t), y = 0\\ t \in [1, 2): x=0, y = \tan(\pi(1-t)) \end{cases}$
For $t=\frac 1 2$ the point $(\pm \infty, 0)$ is the point at infinity that closes line $y=0$ in $\mathbb P^2$. And similarly for $t=\frac 3 2$ and line $x=0$.
Hence this parameterization of the two lines makes them into a closed curve, with one double point obtained for $t=0$ and $t=1$.

The right-angle turn at $(0,0)$ is because we want to connect the infinity branches that are on the same line, i.e. $(0, +\infty)$ with $(0, -\infty)$, and $(+\infty, 0)$ with $(-\infty, 0)$. For $2$ reasons:

• In $\mathbb P^2$, $(0, +\infty)$ and $(0, -\infty)$ exist and are the same point; idem for $(+\infty, 0)$ and $(-\infty, 0)$.
• This is the way hyperbolas behave.

Another option would be to connect $(+\infty, 0)$ to $(0, -\infty)$, and $(0, +\infty)$ to $(-\infty, 0)$. That would be possible if we close the plan $\mathbb R^2$ not with one point at infinity for each parallel set of lines, like in $\mathbb P^2$, but by only one point at infinity, which would topologically be a sphere $S^2$. The two lines would be a closed curve, but with two crossings, and conics do not have crossings, so this looks less natural.

Then we can take a circle $x^2+y^2=1$ centered on the origin, and parameterize it on $t \in [0,2)$ with $x = \cos (\pi t), y = \sin (\pi t)$.

This allows a bijection between the parameterization of the two lines, and the parameterization of the circle. This is not a bijection between the two lines and the circle, as the two lines have a double point at the origin, which correspond to two opposite points $(0, 1)$ and $(0, -1)$ on the circle.

This relation can now be made a relation between a cone and two planes, by adding a third dimension $z \in (-\infty, +\infty)$ which acts as a second parameter:

Two planes $y=0$ and $x=0$:
$\begin{cases} t \in [0, 1): x=z\tan(\pi t), y = 0\\ t \in [1, 2): x=0, y = z\tan(\pi(1-t)) \end{cases}$

The cone $x^2+y^2=z^2$:
$x = z\cos(\pi t), y = z\sin(\pi t)$

However it would be more fun to have a continuous mapping between the two lines and the circle, using a third parameter, with intermediate values being ellipses and hyperbolas, so I am now working on that.

Answer 2

A definition or generalization of a Conical Sheet can be put into a proper perspective to be among a group of surfaces that concurrent straight line generators create by motion of a director line through a single concurrent point $O$ in 3-space.

The director line $ ABC..$ is an arbitrary curve. It need not be part of any bigger curve or tangential to any surface.

There is an infinite number of planes. We only view them pair wise with point of concurrency $O$. Each plane has parallel segments $(AB, ab), (BC, bc),....$ held in an infinity of similar triangles.

The sheet embeds into 3-space, can be flattened out as zero (Gauss) curvature surface. This is referred to as Bending caused by Isometric mapping During such isometric mapping the angle between a pair of planes changes, that is normal curvature changes. However curvature of lines such as $ AB, bc, CA,.." does not change

In particular if the origin is the concurrent point they can be typically described by a homogenous equation as special cases of conicoid

$$ ax^2+by^2+cz^2 =0 \text { or }~\left(\frac{x}{a}\right)^2 \pm \left(\frac{y}{b}\right)^2 +\left(\frac{z}{c}\right)^2 = 0 $$

If one Tangent plane you are referring to at $(x_1, y_1, z_1 )$ should have an equation

$$ a x x_1+ bbb_1+c zz=0$$

If a plane

$$ px+qy+rz=0 $$

touches the cone then we should have

$$\left(\frac{p^2}{a}\right) + \left(\frac{q^2}{b}\right) +\left(\frac{r^2}{c}\right) = 0. $$

etc.