MEI Conference 2014
Conic sections by
paper folding and
GeoGebra
Tom Button
tom.button@mei.org.uk
Conic sections by paper folding and GeoGebra
Parabola
Start with an A4 sheet of paper.
1. Fold it in half long-ways (to create the y-axis) and draw this in with a pen.
2. Mark a point on the fold a couple of inches from the bottom edge.
3. Fold the bottom edge so that it goes through the point and is perpendicular to the
vertical fold (to create the x-axis) and mark this with a pen.
4. Make repeated folds so that the bottom edge goes through the point – these can be
at any angle.
The envelope of lines created should form a parabola.
Creating the envelope of lines on GeoGebra:
1. Add a point, A, on the y-axis.
2. Reflect the point A in the x-axis to create the line A’.
Tablet app: Enter Reflect[A,y=0] in the input bar.
3. Draw a perpendicular line through A’, perpendicular to the y-axis.
4. Add a point, B, on the perpendicular line.
5. Find the perpendicular bisector between A and B.
6. Switch the trace on for the perpendicular bisector.
Computer software: Right-click on the line and select Trace On.
Tablet app: Enter SetTrace[b,true] in the input bar.
Moving the point B will create the envelope of lines.
In terms of B, which point on each perpendicular bisector contributes to the parabola?
(The point vertically above B).
Creating the parabola as a locus on GeoGebra:
1. Switch the trace off for the perpendicular bisector.
Computer software: Right-click on the line and deselect Trace On.
Tablet app: Enter SetTrace[b,false] in the input bar.
2. Clear all traces by zooming in/out or using Ctrl-f.
3. Create a vertical line through B using the perpendicular line tool.
4. Find the point of intersection, C, of the vertical line and the perpendicular bisector.
5. Use the locus tool to draw the locus of C as B moves along the horizontal line.
Using the properties that C is vertically above B and AC=AB it is possible to derive the
equation of the curve: x² = 4ay.
Ellipse
Start with a circular sheet of paper
1. Mark a point near the circumference.
2. Make repeated folds so that points on the circumference go through the point.
The envelope of lines created should form an ellipse. The second focus of the ellipse is at
the centre of the circle.
Creating the envelope of lines on GeoGebra:
1. Add a point, A.
2. Add a circle with centre A and radius 3.
3. Add a point, B, inside the circle.
4. Add a point, C, on the edge of the circle.
5. Find the perpendicular bisector between B and C.
6. Switch the trace on for the perpendicular bisector.
Computer software: Right-click on the line and select Trace On.
Tablet app: Enter SetTrace[a,true] in the input bar.
Moving the point C will create the envelope of lines.
In terms of C, which point on each perpendicular bisector contributes to the parabola?
(The point on the line joining C to the centre).
Creating the ellipse as a locus on GeoGebra:
1. Switch the trace off for the perpendicular bisector.
Computer software: Right-click on the line and deselect Trace On.
Tablet app: Enter SetTrace[a,false] in the input bar.
2. Clear all traces by zooming in/out or using Ctrl-f.
3. Create a line through A and C.
4. Find the point of intersection, D, the line through AC and the perpendicular bisector.
5. Use the locus tool to draw the locus of D as C moves along the horizontal line.
Using the property that DC=DB and D lies on the radius it is possible to derive that the locus
is the set of points such that the sum of AD and DB is constant.
The ordering activity shows how to derive:
x2 y 2

 1.
a 2 b2
Moving B outside the circle gives a hyperbola for the locus.
Equation of an Ellipse – Sorting Activity
x2  c2  y 2  a2 
c2 x2
a2
x 2  2cx  c 2  y 2  2a  x 2  2cx  c 2  y 2
 a2  c2  2
2
2
2

x  y  a c
2
 a

x 2  2cx  c 2  y 2  a 
cx
a
( x  c ) 2  y 2  ( x  c ) 2  y 2  2a
x 2  2cx  c 2  y 2  a 2  2cx 
c2 x2
a2
x 2  2cx  c 2  y 2  4a 2  4a x 2  2cx  c 2  y 2  x 2  2cx  c 2  y 2
4a x 2  2cx  c 2  y 2  4a 2  4cx
x2
y2

1
a2 a2  c2
Equation of an Ellipse – Solution
( x  c ) 2  y 2  ( x  c ) 2  y 2  2a
x 2  2cx  c 2  y 2  2a  x 2  2cx  c 2  y 2
x 2  2cx  c 2  y 2  4a 2  4a x 2  2cx  c 2  y 2  x 2  2cx  c 2  y 2
4a x 2  2cx  c 2  y 2  4a 2  4cx
x 2  2cx  c 2  y 2  a 
cx
a
x 2  2cx  c 2  y 2  a 2  2cx 
x2  c2  y 2  a2 
c2 x2
a2
c2 x2
a2
 a2  c2  2
2
2
2

x  y  a c
2
a


x2
y2

1
a2 a2  c2