Conic Sections

A conic section is the curve that follows the intersection of a cone (or double cone) and a plane. Objects whose orbits are well explained by Newtons law of gravitation (i.e. everything for which GR isn’t needed) follow conic sections. Which type of conic section they follow depends on their orbital parameters.

Here we visualize the intersection of a cone and a plane and briefly run through some the math of how it works.

The Math

A plane is defined as the set of all points that satisfy,

$$Ax + By + Cz = D$$

Where $A$ , $B$ , $C$ , and $D$ are constants (e.g. WolframAlpha). The plane above is defined by,

$$z = 1$$

A cone with the vertex at the origin and with an opening surrounding the $z$ axis is defined by,

$$x^2 + y^2 = \tan^2(\theta) z^2$$

Where $\theta$ is the opening angle (e.g. WolframAlpha). If we then rotate the cone through some angle $\phi$ around the $x$ axis (which is toward us in the diagram above), we can apply the following rotation matrix,

$$\begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos(\phi) & -\sin(\phi) \\ 0 & \sin(\phi) & \cos(\phi) \end{pmatrix}$$

and therefore,

$$x^2 + (\cos(\phi)y + \sin(\phi) z)^2 = \tan^2(\theta) (\cos(\phi) z - \sin(\phi) y)^2$$

(e.g. WolframAlpha).

We can solve for the intersection of these surfaces by substituting $z = 1$ into the definition of the cone,

$$x^2 + (\cos(\phi)y + \sin(\phi))^2 = \tan^2(\theta) (\cos(\phi) - \sin(\phi) y)^2$$

which gives us a function of $x$ and $y$ . The intersection is the set of points that satisfy this equation. Unfortunately this solution is a bit messy, but it can be written¹

$$x^2 = a y^2 + b y + c$$ (1)

where

$$\begin{align} a &= \tan^2(\theta) \sin^2(\phi) + \sin^2(\phi) - 1 \\ b &= -2 \tan^2(\theta) \sin(\phi) \cos(\phi) - 2 \sin(\phi) \cos(\phi) \\ c &= -\tan^2(\theta) \sin^2(\phi) - \sin^2(\phi) + \tan^2(\theta) \end{align}$$ (2)

This solution results in three different shapes depending on the choices of $\phi$ and $\theta$ .

Ellipse

The most general form of an ellipse is,

$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$ (3)

as long as the coefficients satisfy,²

$$B^2 - 4AC < 0$$

The intersection is of the correct form (we can rewrite (1) as $x^2 -ay^2 - b y - c = 0$ ), and so we just need to check the second condition. $B = 0$ (as we have no $xy$ term), and $A = 1$ (the coefficient of $x^2$ ), and so the inequality becomes,

$$\begin{align} -C = a &< 0 \\ \tan^2(\theta) \sin^2(\phi) + \sin^2(\phi) - 1 &< 0 \\ \sin^2(\phi) (\tan^2(\theta) + 1) &< 1 \\ \sin^2(\phi) &< \cos^2(\theta) \end{align}$$

What does this tell us? Well, for small opening angles, $\cos^2(\theta)$ is large and so the inequality is almost always satisfied. When is it not? When $\phi \approx \pi / 2$ , specifically when $|\phi - \pi / 2| < \theta$ .³

When the cone is opening up along the plane we do not form an ellipse. How close to the plane this happens depends on the opening angle.

Hyperbola

What happens in the region where the ellipse is invalid? It becomes a hyperbola, the most general form of which is the same as an ellipse,

$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$

But has the opposite requirement,

$$B^2 - 4AC > 0$$

Following the same argument as before the intersection is a hyperbola when, $|\phi - \pi / 2| > \theta$ .

Parabola

So we have an ellipse when $\theta + \phi < \frac{\pi}{2}$ , and a hyperbola when the sum of the angles is $> \frac{\pi}{2}$ . What happens at the crossover point when one arm of the cone is parallel to the plane at $\theta + \phi = \pi / 2$ ? In that case, $\tan(\theta) = \cot(\phi) = \frac{\cos(\phi)}{\sin(\phi)}$ . Substituting this into (2) we get,

Which gives us a parabola after we substitute into (1) and solve for $y$ .