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.
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.