How do I calculate the curve of a parabolic mirror to melt the ice from my street gutter?

I want to make a concave parabolic mirror that will have a focal point six inches below where I place it (on the edge of the sidewalk) so that it will melt the ice collecting in the gutter below and clear a path for the water to drain. I will probably be using shiny stainless steel for the reflector, so it won't be a fully reflective mirror. What would be the ideal size, and how do I calculate the curve needed to concentrate the heat to that focal point?

Edit: Here's a real life example of someone doing this unintentionally: Reflected light from London skyscraper melts car

• the focal point would be between the sun and the mirror so mirrors would be a poor choice for heating things on the ground. you should try lenses. checkout GREENPOWERSCIENCE on YouTube. you can make a huge lens with water pooling in a sagging stretched plastic plane – gregsan Dec 29 '13 at 7:53
• I edited my post with a link to a real world example. In higher latitudes the sun comes across the sky at a lower azimuth in the sky (from the south) so a refractive lens would probably not be ideal. This wouldn't be a simple parabola that points directly at the sun. – Nathan L Dec 29 '13 at 23:49
• @gregsan: I'm not sure if this is what the OP meant, but you don't need a complete parabola. – lionelbrits Dec 29 '13 at 23:56

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)