# Equivalence of Definitions of Conic Section

## Theorem

The following definitions of the concept of **Conic Section** are equivalent:

### Intersection of Plane with Cone

Let $C$ be a double napped right circular cone whose base is $B$.

Let $\theta$ be half the opening angle of $C$.

That is, let $\theta$ be the angle between the axis of $C$ and a generatrix of $C$.

Let a plane $D$ intersect $C$.

Let $\phi$ be the inclination of $D$ to the axis of $C$.

Let $K$ be the set of points which forms the intersection of $C$ with $D$.

Then $K$ is a **conic section**, whose nature depends on $\phi$.

### Focus-Directrix Property

A **conic section** is a plane curve which can be specified in terms of:

- a given straight line $D$ known as the directrix
- a given point $F$ known as a focus
- a given constant $\epsilon$ known as the eccentricity.

Let $K$ be the locus of points $b$ such that the distance $p$ from $b$ to $D$ and the distance $q$ from $b$ to $F$ are related by the condition:

- $(1): \quad q = \epsilon \, p$

Then $K$ is a **conic section**.

Equation $(1)$ is known as the **focus-directrix property** of $K$.

## Proof

Consider the cone in which $\alpha$ is half the opening angle.

Let the tilting angle of the slicing plane to the horizontal be $\beta$.

The conic section thus defined is given here as an ellipse, but the argument holds for the other cases also.

Inside the cone, let a sphere be inscribed which is tangent to the slicing plane at the point $F$.

Let the sphere be tangent to the cone along the circle $C$.

Let $d$ be the line in which the slicing plane intersects the plane of the circle $C$.

It is to be demonstrated that the conic section induced by the slicing plane has $F$ as its focus and $d$ as its directrix.

Let $P$ be a point on the conic section.

Let $PQ$ be constructed parallel to the axis of the cone such that $Q$ lies on the plane of the circle $C$.

Let $R$ be the point where the generatrix of the cone through $P$ intersects $C$.

Let a perpendicular be constructed from $P$ to $d$, intersecting $d$ at $D$.

Hence:

- $\beta = \angle PDQ$

We have that $PR$ and $PF$ are both tangents to the sphere from the same point $P$.

Thus $PF$ and $PR$ are the same length:

- $(1): \quad PR = PF$

From the right triangle $\triangle PQR$:

- $PQ = PR \cos \alpha$

From the right triangle $\triangle PQD$:

- $PQ = PD \sin \beta$

Thus:

\(\ds PR \cos \alpha\) | \(=\) | \(\ds PD \sin \beta\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds \frac {PR} {PD}\) | \(=\) | \(\ds \frac {\sin \beta} {\cos \alpha}\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds \frac {PF} {PD}\) | \(=\) | \(\ds \frac {\sin \beta} {\cos \alpha}\) | from $(1)$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \frac {PF} {PD}\) | \(=\) | \(\ds \frac {\cos \gamma} {\cos \alpha}\) | where $\gamma = \angle DPQ$ |

Let us define:

- $e = \dfrac {\cos \gamma} {\cos \alpha}$

For a given slicing plane and cone this is constant.

It is noted that:

- $(1): \quad$ when $\gamma < \alpha$, that is, the tilting angle is less than half the opening angle of the cone, the conic section is an ellipse

- $(2): \quad$ when $\gamma = \alpha$, that is, the tilting angle equals half the opening angle of the cone, the conic section is a parabola

- $(3): \quad$ when $\gamma > \alpha$, that is, the tilting angle is greater than half the opening angle of the cone, the conic section is a hyperbola

- $(4): \quad$ when $\gamma = 0$, that is, the slicing plane is parallel to the plane of the circle $C$, the conic section is a circle and there is no directrix

The result follows.

$\blacksquare$

## Sources

- 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {A}.8$: Pappus (fourth century A.D.): Appendix: The Focus-Directrix-Eccentricity Definitions of the Conic Sections