## Monday, September 21, 2009

### More on parabolas

Let's continue with the previous problem. We have the parabola
 y = x2y' = 2x

The slope at any point is 2x. To find the focus, we require the y-value where the slope equals 1, that occurs when
 x = 1/2, y = 1/4

So the focus is at the point:
 x = 0, y = 1/4

Now, we want to consider all points x. At every such point
 y = x2

The line from the focus (superimposed in magenta on our old diagram) has the slope at every point of:

 (x2 - 1/4) / x

Rather than deal explicitly with calculating angles, I'm going to use vectors. There are three of them in this problem. It doesn't matter whether we take the vectors in their forward or reverse directions, so I'll consider IN as

the reverse of the unit vector of light coming in:

 IN = [0,1]

The reverse of the vector of light coming out calculated above, and the slope are:

 OUT = [x, x2 - 1/4]M = [1, 2x]

What we are trying to prove is that the angle between OUT (magenta arrow) and M (slope or tangent to the curve at the magenta point) is the same as the angle between IN (blue dotted line) and M.

In vector algebra, the dot product

 A • B = Σ Ax Bx + Ay By = |A| |B| cos θ

where the absolute value of the vectors comes from Pythagoras, and θ is the angle between them. In the language of vectors, we claim that the angles for the two vector comparisons are the same, and therefore:

 IN • M / |IN| |M| = OUT • M / |OUT| |M|IN • M / |IN| = OUT • M / |OUT|IN • M = 0 * 1 + 1 * 2x = 2x|IN| = 1OUT • M = x + 2x * (x2 - 1/4)|OUT| = √ [x2 + (x2 - 1/4)2 ]

So the left-hand side is just 2x.

The right-hand side looks complicated but it is easily simplified. The numerator is:

 x + 2x * (x2 - 1/4) = 2x (1/2 + x2 - 1/4) = 2x (x2 + 1/4)

The denominator is:

 √ [x2 + (x2 - 1/4)2 ] = √ [x2 + x4 -1/2 x2 + 1/16 ] = √ [x4 + 1/2 x2 + 1/16 ] = √ [(x2 + 1/4)2 ] = x2 + 1/4

Cancel x2 + 1/4 from both top and bottom, leaving 2x, and the identity is verified! There is one more question, however, and that is whether we can generalize this result to any parabola. That will have to wait.