Chapter 13: Conic Sections

Chapter 13: Conic Sections

Parabolas through Paper Folding

Take a sheet of tracing paper or wax paper and draw a line segment across its width about 5 centimeters from one end -- this will be the directrix of your parabola. Mark a point near the center of your sheet of paper -- this will be the focus of your parabola. Now fold the paper so that the directrix passes over the focus and make a good crease. Repeat this action with many different folds so that different positions along the directrix coincide with the focus point. (One way to do this is to start by folding one end of the directrix over to the focus point and gradually working your way along the directrix to the other end.) Hold your unfolded paper up to a light source.

- What shape does the envelope of folds create?

- What is the connection between the crease lines and the focus point and directrix?

- What has this to do with the Focus-directrix definition of a parabola?

Activity 13.1

Think about how you might construct the fold line starting with a fixed point and fixed line in GSP. As you want the point to fold onto the line in many different positions, place a free point on the line. This free point on the line will be the position of your folded point. Look back at your paper folding. How does the fold line relate to the fixed point and the point at which this point touches the line when folded? Join the two points with a segment on your paper. How does your fold line relate to this segment? Where does it intersect the segment?

From the above questions you probably found that the fold line intersects the segment at its midpoint and that the fold line is perpendicular to the segment. The fold line is the perpendicular bi-sector of the segment joining the fixed point (focus) to its folded image on the fixed line (directrix). Construct this perpendicular bisector in your GSP sketch using the following steps:

1. In a new sketch draw a line (directrix) and a point (focus) not on the line. Place a free point on the line and label it “Animate me”.
2. Construct the segment between point “focus” and the point “Animate me”.
3. Construct the midpoint of this segment.
4. Select the midpoint and the segment and choose Perpendicular Line from the Construct menu.

You can now move the point “Animate me” on the directrix to see the fold line change. Select the fold line (perpendicular bisector) and turn on the Trace function by going to the Display menu and choosing Trace. Now move point “Animate me” again. You should see the traces of your fold lines appear like the folds on your paper. You can obtain a more evenly spaced set of traces by using an Animate Button. With the point “Animate me” and the directrix line selected, go to the Edit menu and choose Action Button-Animation. Accept the default settings. A button should appear in your sketch. Click on your button to trace out your parabola. Clicking the button a second time will stop the animation.

Change the position of your focus point and animate again. Change the orientation of your directrix line and reanimate. It is also possible to construct these traced lines by constructing the locus of the perpendicular bisector as point “Animate me” moves on the directrix. First turn off the trace of the perpendicular bisector by selecting the line and going back to the Display menu, then choosing the checked item Trace (this will turn off the trace function). You will also need to erase the traces you have made by choosing Erase traces from the Display Menu. Now select only the perpendicular bisector and the point “Animate me”. Go to the Construct menu and choose Locus. All the fold-lines should reappear as in the sketch in figure 3.8. This locus of the fold lines is called an “envelope.” It is this envelope that produces the parabolic curve. Each fold line is, in fact, a tangent to the parabola.

Assignment 13.1: Construct the point on the perpendicular bisector that will be the point of tangency for the parabola. You can then use this point to construct the parabola itself, rather than its envelope. Remember the locus definition for the parabola. The point on the parabola must be equi-distant from the focus and the directrix. Figure 13.1 provides a big hint for constructing this point!

Figure 13.1: Traced Locus of the Perpendicular Bisector

When you have constructed the point on the perpendicular bisector that will also be on the parabola, select this point and the point “Animate me” and choose Locus from the Construct menu. The actual parabolic curve should appear along the boundary of the envelope of fold lines. You can hide or delete the locus of fold lines so as to see just the curve itself. Move your focus point closer or further away from the directrix. How does this affect the shape of the parabola?

Paper folding and Other Conics

Instead of folding a line across your paper onto a point you can fold a circle onto the point. (It is actually easier to fold the point onto a circle.) Move the point around the circle, making a new fold line for each new position. What shape does the envelope of folds create? If you used a point outside the circle try the same thing (with a new sheet of paper) with a point inside the circle (and vice-versa). Simulate this activity with the GSP.

Exploring Parabolas and Orthocenters

The locus of the Orthocenter, as a vertex of the triangle moves on a segment parallel to the base of the triangle, appears to be shaped like a parabola. In order to explore this possibility we shall investigate various ways of constructing parabolas with GSP and paper folding. The locus definition of a parabola is “the path of a point that moves so that it is always equidistant from a given fixed point and a fixed line.” The fixed point is called the focus of the parabola and the fixed line is called the directrix.

Locus of the Orthocenter of a Triangle

Return to your animation of the Orthocenter of a triangle. As the vertex of the triangle is animated on a segment parallel to the base of the triangle, the trace of the Orthocenter looks like a parabola. If it is a parabola it ought to be possible to find its focus and directrix.

Assignment 13.2: Construct the Focus and Directrix of the locus of the Orthocenters of a family of triangles with same base and height.

The Parabola Tracer construction may suggest a way to construct the focus and directrix. Return to your parabola sketch. Look at the constructed locus of the point on the parabola. Move point “Animate me” to see how the position of this point moves around the parabola. Now run the animation of your Orthocenter.

Big Hint: Where is the Vertex of the orthocenter parabola? Construct it. Do the end points of the base of your triangle lie on the parabola?

[One solution to this challenge can be found in the file Focus_Directrix_of_OC.gsp on the accompanying CD.]

CONIC EQUATIONS

The general equation for a circle centered at the origin with radius r is x2+y2=r2. Think about extending this equation to a circle at center (h, k) by remembering the vertex form of the quadratic function.

The general form of the quadratic, of course, gives you the parabola. You could investigate the focus-directrix form of the quadratic equation, given a focus at point (h, f) and a directrix at y=d (you might want to start with focus at (0, f). Use what you now know about the relationship between any point (x, y) on the parabola and its distance to the focus and the directrix.

The general form of the equation for an ellipse is very similar to the equation of a circle except that we now have major and minor axes rather than one radius. See if you can find out what the general equation is for the ellipse based on the values for half the major (a) and half the minor (b) axes of the ellipse. You can start with an ellipse whose major and minor axes lie on the x and y axes and centered at the origin (similar to a circle centered at the origin).

The simplest form of an equation for the hyperbola is a rational function. The function f(x)=1/x gives a "square hyperbola" centered at the origin. You can generalize this function to give a square hyperbola centered at any point.

Here is a general form of an equation that can be used to form ANY conic:

ax2 + by2 + cxy + dx + ey + f = 0

See if you can find information on this form, and the special values of the SIX coefficients (a, b, c, d, e, f) that produce each specific conic!