# How does the eccentricity of a conic section define its shape?

Problem:

Let $P$ be a point in the plane, $L$ a line containing $P$, and $\varepsilon$ a positive number.

The triple $(\varepsilon, L, P)$ will then define a degenerate conic section. $\varepsilon$ dictates what type this degenerate conic section is. How?

Hint: One can assume that $L$ is the line $x=0$ and $P = (0,0)$.

My attempt:

...has so far been limited to googling, but any resource I find, simply states what I already know; that the eccentricity does in fact dictate shape.

I've tried looking at the formulae for the various shapes, but no luck.

• Well eccentricity does dictate shape of a conic. $0<e<1$ denotes an ellipse ($e=0$ circle), $e=1$ parabola, and $e>1$ hyperbola. Is this what you are asking? – dardeshna Jan 16 '15 at 5:49
• @dardeshna - No, that's the premise, not the question. The title asks "how", and I've also bolded this in the question itself. – Alec Jan 16 '15 at 7:32

The focus-directrix definition of a conic section (which seems to be the definition that you're referring to) is as follows:

Given a line $L$, a point $P$, and a real number $\varepsilon>0$, a conic section is the locus of all points such that the distance to $P$ is $\varepsilon$ times the perpendicular distance to $L$.

$P$ is referred to as the focus, $L$ as the directrix, and that number $\varepsilon$ is called the eccentricity. Note that we cannot get a circle out of this definition, as letting $\varepsilon=0$ would give us just the point $P$ and nothing else. Now, if the point $P$ does not lie on $L$, we get the wonderful ellipses and parabolas and hyperbolas which you can study to your heart's content. However, if the point $P$ lies on $L$, we get the exceptionally boring degenerate conics.

So what do the degenerate conics look like? We will, as you suggest, assume the focus is $(0,0)$ and that $L$ is the line $x=0$. Suppose firstly that $\varepsilon<1$. A point $(x,y)$ is in the conic section if and only if the distance to $P$, or $\sqrt{x^{2}+y^{2}}$ is equal to $\varepsilon x$ ($x$ is the perpendicular distance to the line $x=0$). This would imply that the distance to $P$ is less than the distance to $L$ which is clearly impossible, since the hypotenuse must always be longer than the leg of a right triangle. Unless, of course, $(x,y)=(0,0)$. So if $\varepsilon<1$, we just get a single point.

What if $\varepsilon =1$? Well, it shouldn't take too much imagination for you to prove that the only points where the distance to the origin is equal to the distance to the y-axis are points of the form $(x,0)$.

And what if $\varepsilon>1$? Remember that $\varepsilon$ is the distance to $P$ divided by the distance to $L$, so for $(x,y)$ on the conic, we have

$$\varepsilon=\frac{\sqrt{x^{2}+y^{2}}}{x}$$

by a simple bit of rearrangement, we get that

$$y=(±\sqrt{\varepsilon^{2}-1})x$$

Since $\varepsilon>1$ we have that $\sqrt{\varepsilon^{2}-1}>0$. So this last case would be a pair of lines intersecting at $(0,0)$ with gradients of equal absolute value and opposite sign.

Now of course, all these conic sections are just the boring ones with the focus on the directrix. To get the others, you need to look at the more ordinary case where the focus is not on the directrix.

The eccentricity of a conic can be defined as the distance between the foci divided by the distance between the points of intersection of the conic with its major axis (its ends).

In a circle, the foci are coincident at the center of the circle. Thus, $\epsilon=0$.

In an ellipse, the foci are distinct and inside the ellipse and the ends are the points which are the farthest apart on the ellipse, so $0\lt\epsilon\lt1$.

A parabola is a stretched out ellipse with one focus and end at an infinite distance. The difference between the distances whose ratio defines the eccentricity is twice the focal length of the parabola, yet both have passed to infinity. Thus, $\epsilon=1$.

The foci of a hyperbola are inside alternate lobes of the hyperbola, and the ends are the points of the two lobes which are closest together. Thus, $\epsilon\gt1$.

Since the eccentricity is given, the shape of the conic section is determined. The degeneracy must be in the distance between a focus and an end of the conic section. In this answer, it is mentioned that a circle with infinite radius can be considered to be a line. However, since we are given a point on a line, the degeneracy is probably of small distance between a focus and an end. In that case we have the following.

A degenerate circle or ellipse is a point.

A degenerate parabola is a ray.

A degenerate hyperbola is a pair of intersecting lines.

• Good, but these are all non-degenerate cases of conics. In the question, I explicitly ponder the degenerate cases. – Alec Jan 16 '15 at 10:30
• Ah. Do you mean something like: a degenerate circle is a point, a degenerate ellipse is a point or a line segment, a degenerate parabola is a ray, and a degenerate hyperbola is two intersecting lines? It is hard to tell from the question how $\epsilon$, $L$, and $P$ determine the conic. – robjohn Jan 16 '15 at 10:33
• Exactly. I'm not familiar with the "ray" terminology in this case, but I assume you mean two parallel lines. But yes, I'm trying to figure out how $\varepsilon$ triggers these different cases. – Alec Jan 16 '15 at 10:35
• Here is Wikipedia's definition of a Ray. – robjohn Jan 16 '15 at 10:37