I hope I am not bothering you in any way, but there is a question that has been bothering me for awhile now. I have asked my teacher but he does not know the answer. I have been researching it online and in some text books for about a month now and have come up with nothing. I do not want this to seem like my math teacher knows nothing, he is a very intelligent man.
I have taken Calculus 1 and 2 in high school and am leaving for college to study engineering in the fall. The question I have which I would appreciate any little bit of information or direction on is: Where does the 4 come from in the parabola equation:
(x – h) 2 = 4 p (y–k) ?
This has been bothering me simply because I wish to know the why and how of things. I am sorry if I am taking up you time and I appreciate you for reading my question. Thank you.
You realize of course that the textbook could easily get rid of the factor 4, by simply introducing a new constant q = 4p, which leads to the same formula but without the puzzling factor. The fact that the "4" is included suggests that the constant "p" has some special significance, which would be lost if the equation were written otherwise.
This indeed is the case.
When you define a parabola by its equation
(x – h)2 = 4 p (y–k)
you are using coordinate geometry, invented by Descartes in the 1600s. Today we prefer such a definition, because it allows geometrical properties to be handled by means of numbers, and our age is very, very good in handling numbers.
The ancient Greeks, the first to study parabolas, were not so good with numbers (maybe because they lacked the decimal system), but they were very good at pure geometry. They defined a parabola as the collection of points, each of which had the same distance -- call it "q"--from some point F, its focus, as it had from a straight line, which they called the "directrix" of the parabola.
Let us try to translate this condition into the language of coordinate geometry. Get some paper, draw on it (x,y) axes, and add a freehand sketch of the parabola y = x2 , passing through the origin and rising above it symmetrically, on both sides of the y-axis. You need not number the axes. After that please follow one by one the steps outlined below, on paper. Do not try to skip anything, and proceed past any point only after everything preceding is clear. Then when you are done, explain it all to your math teacher.
On the graph you drew above, mark a point D= (0, –q) a distance q below the origin. Place D about as far from the origin as 1/4 of the height of the y axis you have drawn rising above it. Draw through D a straight line parallel to the x-axis (its equation is y=–q). That will be the directrix.
By symmetry, you expect the focus F to be on the y-axis (only then does the Greek condition produce a curve symmetric with respect to that axis). The origin is one of the points of the parabola, at a distance q from the directrix, so by the Greek condition, the focus F is a distance q on the other side, at the point (0, q).
Next, mark some point P = (x,y) on your parabola; to best illustrate the argument, choose it with y of about 2q or 3q. Also draw a perpendicular line from P to the directrix, a line parallel to the y-axis. Say it meets the directrix at point Q=(x, –q).
You have drawn PQ, now add one last line PF. By the Greek definition, they are equal in length
PF = PQ
PQ = y + q
(PF)2 = (y+q)2
Expressing (PF) 2 by the theorem of Pythagoras
(y–q) 2 + x2 = (y+q) 2
Multiply out the squares! You find you can now subtract from both sides y2 and also q2, leaving just
–2yq + x2 = 2yq
x2 = 4qy
But suppose we are working, not in (x,y) coordinates where the lowest point of the parabola happens to strike the origin, but in (X,Y) coordinates, at which the lowest point is at some arbitrary point (X,Y) = (h,k) . The connection between the two systems is a simple shift in x and y :
x = (X–h) y = (Y–k)
The equation of the parabola now becomes
(X–h)2 = 4q(Y–k)
That is of course your formula. You can see that when the factor 4 is included, then q (or p in your version) has a geometrical meaning, it is the equal distance which all points on the parabola maintain, from a point and from a straight line.