Back to Analytic Geometry
Analytic Geometry:
AG 009 Parabola:
By William E. Steinman:
Let’s look at the conic section we call a parabola.
We define a parabola as the locus of all points that are equidistant from a fixed point and a fixed line. The distances are the undirected distances.
The fixed point is called the focus of the conic and the fixed line is the directrix. and the constant ratio the eccentricity, which is represented by a lowercase e.
Huh?
Let’s take an example to try to make this clear. Start with our good old standard two dimensional grid.
Assume that F on the x axis with coordinates of a, 0 is the fixed point, the focus.
Assume that the fixed line, the directrix is the same distance from the Y axis in the negative direction.
Now make a leap and assume this sketch is an accurate rendition of a parabola. An artist I am not!
If it were accurate, the ratio of the distance (M,P) to the distance (F,P) would always be the same or constant.
That would be true if the point were above or below the X axis.
Swell!
We have previously learned that the general form of an equation in the second degree is Ax2 + Bxy + Cy2 + Ey + F = 0.
The equation for a parabola is a special case of that equation.
One form of the equation for the parabola is y = ax2 + bx + c.
The particular equation will depend on where we place the focus and the fixed line or directrix.
If we choose our axis, as we did above we will get the simplest equation. That is when the y axis splits the distance between the directrix and the focus.
If we let p = the directed distance from the focus to the directrix, the coordinates of F will be (1/2p, 0).
The equation for the directrix will be x = -1/2p.
We can let d1 and d2 be the undirected lengths of the lines FP and MP.
For every point where d1 = d2 it is clear that d12 will equal d22
If we simplify this we will end up with the equation y2 = 2px and this is the simplest equation we can obtain for a parabola.
That is, every point in the plane that has d1 = d2 will satisfy our equation. The clinker in this is we have not proved the converse, that there are no other points that satisfy the equation.
Let us do that.
Any point (x, y) that satisfies y2 = 2px will be such that
Y2 + (x – p/2)2 = 2px + (x – p/2)2
All we did here was add the same value to both sides of our equation.
That is a very legitimate operation that we learned way back in algebra.
We can diddle the result like this
Y2 + (x – p/2)2 = 2px + x2 – px + p2/4
Y2 + (x – p/2)2 = x2 + px + p2/4
Y2 + (x – p/2)2 = (x + p/2)2
This tells us that d12 = d22
Look at our original d1, d2 equations above.
Or, to belabor the obvious, if we take the root of both sides we will get d1 = d2
If you think it through, this proves that that all points where d1 = d2 are on this curve.
Conversely, there are no points on this curve where d1 ≠ d2
That’s enough for now.
We can finish with the parabola next time.