I was recently re- reading through some old (1920) notes from the Philosophical Magazine that Dave Renfro sent me (THANKS, Dave) and came across a nice problem based on an old principal of parabolas known to Archimedes. The method, I learned in the article, was used by Archimedes in his

*On Floating Bodies, book two*.. in the course of investigating the equilibrium of a floating parabaloid of revolution.. In the article the author gives this theorem and quotes it as if it is well known... Yet it seems not to appear in texts much then (1920) or now. I did find the question in An elementary treatise on pure geometry:**with numerous example by John Wellesley Russell.**Here is the problem: Given a point on the curve A, and the slope of the tangent at that point, (AC) and a second point on the curve B, construct (in the classical sense) additional points on the parabola... C was place above B by chance, and can be anywhere along the tangent. I have placed the problem on a coordinate grid to present it as a function of x, but the actual coordinates of the points have no influence on the construction, although it is assumed that you know the direction of the axis of symmetry (in this case, vertical).

The calculus student in you might want to attack this analytically, but time for that later. Let me show you the Geometric method of Archimedes.

We begin by constructing a vertical line through B, and selecting a point D, somewhere along this line . Through this point draw another line parallel to the tangent and a second through point A. Finally draw a secant AB of the parabola.

And then the final act. Mark the point where the parallel to the tangent intersects the secant AB. From this point, extend a vertical line to find the point where it intersects AD. This final point P is on the parabola AB with a tangent of AC at A, and it will be for whatever point D you picked originally.

With straight edge and compass, you would have to pick a new point D, then recreate another parallel to the tangent, find another intersection at E, and then vertically transfer that up to the line AD for each new point. But with Geogebra, you can construct, and then just move D and watch P trace out the parabola.

So NOW let’s do a little calculus.

If the original points are at (0,0) [why not] and (p,q) and the slope of the tangent is m, then we need to find A, and B (C=0 by a clever choice of coordinates) for the parabola y= Ax

^{2}+ Bx. We also know that at x= 0, dy/dx = m so 2Ax +B = m so B must be the slope m. Now we just need to fill in y= Ax^{2}+ mx and passing through (p,q). This gives us q = Ap^{2}+mp and we can solve for A = (q-mp)/p^{2}.With my selected easy values of m=1 and (p,q) = (4,1) we see that y= -3x

^{2}/16 + x .**Two more nice problems for Calculus students**that point out things that are easy not to notice in the rush to memorize rules and such..

PROVE each:

1) If you draw to tangents to a parabolic function, the x-coordinates of their intersection is the arithmetic average of the x-coordinates of the two points of tangency.

2) If you draw the tangents to any parabola at the endpoints of the latus-rectum, they will always be perpendicular.