A parabola is a conic section, like circles and ellipses, and all three types of curve can be defined by a focus (or in the case of the ellipse two foci). In the case of a parabola we draw a straight line (the directrix) and choose a point (the focus) and the parabola is the set of points that are an equal distance from the directix and the focus:
(image is from the Wikipedia article linked above).
It should be obvious from the diagram that the focus is at the point where parallel rays hitting a parabolic mirror will converge, so it is both the focus in a mathematical sense and the focal point in an optical sense.
Response to comment:
The centre of curvature is the centre of the osculating circle. We can draw this because near the vertex the parabola looks like part of a circle.
Take the unit circle centred at $(0, 1)$:
$$ (y-1)^2 + x^2 = 1 $$
or expanding this:
$$ y^2 - 2y + 1 + x^2 = 1 $$
If we are very near the origin $y^2 \ll y$ so we can approximate the above expression by:
$$ -2y + 1 + x^2 = 1 $$
which rearranges into the equation for a parabola:
$$ y = \tfrac{1}{2} x^2 $$
And the focus of this parabola is at $(0, \tfrac{1}{2})$. I've used a special case to make the working simple, but you can generalise this to any circle passing centred on the $y$ axis and passing through the origin.
The important thing to notice here is that the layer of black goes behind the Mylar layer. So the light that will be absorbed is whatever penetrated the surface aluminum layer (a small fraction). That is, most of the light will be reflected off the front surface, and only a small amount will be affected by this choice.
Now we ask "Why don't we want that light reflecting?
At this point it helps to have a little understanding of wave optics. The rainbow colors seen on the surface of puddles in the road come from the interference of light reflected from the top of a floating oil layer and that reflected from the transition from oil to water. When you look at a colored band you are seeing the light from some of the visible spectrum that is constructively interfering while light from another part destructively interferes. More interesting still are the black stripes sometimes seen near the edges of the oil slick: in that regime the slick is much thinner than a wavelength of visible and because the front reflection is phase inverting and the back one is not (don't ask, you don't actually care right now) all visible light is destructively interfering in reflection.
The Mylar is designed to be as thin as possible, and the energy from the sun covers a wide spectral band, so in any given direction from the surface some wavelengths will be destructively interfering if reflections from the rear surface are allowed.
You read that right: the "extra" light reflected from the rear surface of the Mylar could result in less energy getting to the collector.
This effect can be even worse if the layer acts as a Fabry–Pérot etalon, because it can magnify the usually small effect of a single reflection interfering (which depends on the amount of light that passes the first surface).
Best Answer
Ice and snow reflect light quite a bit, so you might be better off focusing the light onto something black and then using the radiant heat from that to melt the snow. But in case you want to do it directly, I would suggest pointing the axis of the parabola at the sun. You don't need a complete parabola (whatever that means!), as any rays that hit it parallel to the axis will end up at the focal point.
A parabola with equation $x = \frac{y^2}{4a}$ (opens rightward) has a focus at $x=a$. So assuming the light comes in along the $x$ axis, the ground would some line passing through the point $(x=a,y=0)$ sloping downward. I.e., $y= \tan\theta\, (x-a) $. You only need the part of the parabola that is to the right of that line (i.e., above ground)