Have a look at this construction.

You see:

• the parabola y=x^2

• two points A and B on the x-axis with distance 1

• the two tangent to the parabola belonging to A and B

• the intersection C of these tangents

• the track (green) of C, when A runs on the x-axis

Several questions arise. First of all, where is C relative to A and B? Then, of course, what is that track? Is it really a parabola as it seems.

But before I study this in further detail, let me explain how this construction is done. Since C.a.R. seems to be used more and more as a function plotter, I have chosen this example with the function and its tangents. C.a.R. is a geometry program first of all. But these questions are certainly geometric, especially since the underlying function is a conic section.

• Enable the coordinate grid. This is just for decoration and nothing can be constructed with the grid alone.

• Take the default macro "coordinate grid" and click on (0,0). Alternatively, create a line along the x-axis. Hide the points.

• Create a point A on the x-axis.

• Create the function x^2 with the function tool. Name it "f". Zoom in with the mouse wheel, or the + key.

• Create another point somewhere, and edit it by right clicking it. Fix it to x=x(A) and y=f(x(A)). It will then stay on the parabola, when A moves on the x-axis.

• To create the tangent, we could use a function. However, for speed reasons, we create it as a line. So create another fixed point at x=x(A)+0.1 and y=f(x(A))+diff(f,x(A))*0.1. The diff function numerically computes the derivative of "f" at x(A). instead of 0.1 you can just as well take 1 or any other number. Connect the new point with the point on the parabola by a line. Then hide it.

• Now we make that a macro, so we do not have to repeat the same for B. Set A and f as parameters, and the tangent as a target. Name it "Tangent".

• Create a point somewhere, and fix it to x=x(A)+1 and y=0. Name it B.

• Apply the macro "Tangent" to B and f.

• Intersect the two tangents. Name the intersection point C. Color it green.

• Create a track of C, while A runs on the x-axis. You simply have to choose C and A as parameters for the automatic track tool, since C binds to the x-axis. Click to stop the animation.

• Take the track into the construction by pressing "Ctrl-Enter" (or with the menu item). Color it green.

• I connected A and B with their points on the parabola with a light blue line segment.

You can now move A. Therefore I made A fat. This follows the convention that basic points are fat. Here is the download link for this construction.

Of course, we can study this problem algebraically. In the school, this is obviously the way to go. In fact, it is a nice problem on tangents. Since we want to study more complicated functions later, we use Maple. The problem is simple enough for any algebraic program. The TI89 pocket computer will solve it just as easily.

I am not going to explain Maple here. As a help for starters, f is the formula of the parabola, depending on x, and g is the formula for the tangent in x, depending on t. We use Maple's solver to set two tangents at a and a+1 equal to each other. 

>f:=x^2;
>g:=diff(f,x)*(t-x)+f;
>solve(subs(x=a,g)=subs(x=a+1,g),t);
>subs(x=a,t=%,g);
>subs(a=t-1/2,%);

We get t=a+1/2. So the x-coordinate a+1/2 of the point C is exactly in the middle of A and B. Substituting this result for t into one of the tangent equations, we get the y-coordinate of C as a+a^2. Now we substitute a=t-1/2, and get the point C with coordinates t and t^2-1/4. This is indeed a parabola.

If you wish, you can enter the function x^2-1/4 into C.a.R. to check the result.

Can we prove the result geometrically? It will not be easy. However, the first part is a really nice application of the parabola as the track of all points with equal distance to a given point P and a line g. Have a look at this construction.

 

The line g goes through A and B, and F is the focus of the parabola. Then the two tangents belonging to A and B are the middle perpendiculars on AF and on BF.  Thus FP=PA and FP=PB. We get PA=PB. Thus we have that M is the midpoint of AB.

It is a bit tougher to see that the track is indeed a parabola, in fact the same parabola shifted perpendicular to the basic line g. The following construction is the basis of this proof (download link)

We have to show that SC is constant, when F moves along the perpendicular at D to MD. Given are MB and MD, and of course we assume SM=SF, CB=CF. If you download the construction and move F, the fact that SC is constant is quite surprising. I was able to come up with a computational proof only. One can compute 2*SC*MD=MB^2.

Now let us take another function for our tangent experiment. As soon as we do this, we get rather complicated tracks. On the left, we have x^3 and on the right x^4.

The function for x^3 can be computed with Maple, using the same techniques as above. In the last step, we have solve for a depending on t before we substitute. The function is rather involved.

 

There are two branches, as you can see from the image.  Of course, things are easier, if we represent the track as a parametric curve. We get two parameter functions involving rational functions. I did not try x^4 in Maple. It should be a function, but the equation would be rather complicated.

I have another problem for you.

These are two parabolas with same focus point and two parallel basic lines. On the blue parabola, I fixed a point P, and constructed the two tangents to the green parabola and the tangent to the blue parabola. Proof that the two angles are equal. Here is the construction file.

Consequently, a ray reflected inside the blue parabola will touch the green parabola at several points.

Below is the same result for ellipses with common focal points. Note that both ellipses can also be described as the set of points with equal distance to B and to a circle around A (with different radius, of course).

You can download the construction here. But if you never worked with ellipses given the two focal points, you may wish to figure out how to do that construction yourself.

R. Grothmann