First let us consider the locus of each of the three points
A, B and C, which are particular special cases. The locus of
A is just part of a vertical straight line because A is constrained
to move up and down the door post. The locus of B is the
quadrant of a circle with centre O and radius OB. The locus
of C appears to be a horizontal straight line through O, but
this needs a little more explanation. To show that OC is a
straight line it is necessary to show that angle A OC is a right
0M
0M
0M
angle. Triangles AOB and BOC are both isosceles. If we let
angle A OB be 0 degrees, we can easily show that angle OBC
is 20 degrees, by considering the angles at A and B. It then
follows that angle BOC is 90 - 0 degrees, and so AOC is a
right angle.
O
C
S 90-0
Br
B
r
1A
Interestingly the same argument with an identical diagram
looked at from a different point of view can be used to prove
that the angle in a semicircle is a right angle.
0
A 90-
by Doug French
SB
c
The locus of other points on the door is less obvious.
Examples generated by Cabri are shown for two points, D,
the midpoint of AB, and E, at the foot of the door. This
experimentation suggests that the locus in each case is part
Garage Doors
of an ellipse, but can we prove it?
The way in which 'up and over' garage doors are constructed
and the movement of points on such doors provide some
interesting ideas to explore at various different levels. The
design of my own garage door is illustrated here with a side
view. OX is one of the vertical door posts and CE is the door.
A point A on the door is constrained to slide up and down the
/6
door post, whilst another point B is connected with a rod,
freely pivoted at both ends, to the point O at the top of the
post. The lengths OB, EA, AB and BC are equal and are each
denoted as of length r in the diagram.
EY
O
C
8
Stretched Circles
First let us consider how an ellipse can be defined. The
A
simplest way is to define an ellipse as a stretched circle. The
diagrams show a circle with unit radius, a circle with radius
r and an ellipse with major axis 2a and minor axis 2b, all with
their centres at the origin. The parametric equations in terms
of a parameter 0 are given for each curve.
E
x
This diagram has been drawn using Cabri Geometre (Cabri
Geome'tre, 1996), which enables you to move the door on the
screen and to observe and trace out the locus of various
points. The same thing can be done, although less dramatically, using a model made of Geo-Strips, Meccano or card-
board strips with paper fasteners. The elliptical form of the
locus of various points on the door is the focus of this article.
6
Mathematics
in
x = cos 0
x =rcosO
x = a cos
y= sinO
y = rsin 0
y= bsin0
School,
r=2
March
a=3, b=-2
1999
The larger circle is an enlargement of the unit circle by a
factor of r, which is equivalent to stretching it by a factor of r
in both the x and the y directions. The ellipse can be obtained
by stretching the first circle by a factor of a in the x direction
and a factor of b in the y direction.
By using the identity cos20 + sin20 = 1, the standard
Cartesian equations for the circle and ellipse with centres at
the origin are readily obtained from these parametric equa-
It is worth noting that some designs of'up and over' garage
doors use the 'slipping ladder' principle, rather than have the
door attached to a radial rod from O. The same idea was used
in the past in drawing offices in a device for drawing ellipses
known as the ellipsograph or Trammel of Archimedes. A description of this will be found in Cundy and Rollett's classic
book, Mathematical Models (Cundy and Rollett, 1951).
tions.
Circle centre (0,0) and radius r:
C
x2 + y2 = r2
Ellipse centre (0,0) with major and minor axes 2a and 2b:
B
X2 2
D
a2+b2 =1
O
A
O
It is worth noting in passing that the stretching transfor-
mation also shows why the respective areas are nr2 and nab.
Unfortunately there is not a similar relationship for the
perimeters!
Proving the Locus is an Ellipse
We are now ready to prove why the locus of points on the
door is, in general, an arc of an ellipse. For convenience, the
original diagram has been rotated through 90' anticlockwise
so that OA is taken as the x-axis and OC as the y-axis.
A
Besides the locus it is also interesting to observe the
envelope of lines created by the positions of the door or ladder
as it moves. This is very relevant in the case of a garage door
because a vehicle inside must be below the curve formed by
the envelope if the door is to be opened and closed without
causing any damage!
y
c
,,
B
Pins and String
r/ \D
8
\A
For many people, however, the most familiar way of drawing
an ellipse is the method using two pins and a piece of string
x
O
E
illustrated below. Pins are placed at two points, F, and F,,
Consider first the locus of the point D, the midpoint of
AB. Since triangle OBA is isosceles and BD = DA Ilr,
the
2
and the ends of a length of string are tied to the pins so that
it lies slack between them. The string is then held taut with
a pen which is moved along the string to plot the locus.
x- and y-coordinates of D are:
P
3
x = -r cos0
2
F, F2
1
y = -r sine
2
These are the parametric equations for an ellipse with major
axis 3r and minor axis r.
In a similar way, the locus of E, the point at the bottom
end of the door, is also an ellipse given by the following pair
of parametric equations:
x = 3r cos0
Each point where a pin is placed is a focus of the ellipse.
Varying the positions of the two foci varies the shape of the
ellipse. If they are both at the same point the locus is a circle.
As they are moved apart this becomes a wider and thinner
ellipse, tending towards a straight line when the string is
stretched taut between them. The interesting question now is
to prove that the curve is, in fact, the same curve as the
ellipse that comes from a 'stretched circle'.
It is easy to verify that the length of the string is equal to
the length of the major axis by noting what happens when the
two segments of the string overlap. This can be written as:
y = - r sin0
In this case the major axis is 6r and the minor axis 2r.
Drawing an Ellipse
The locus of points on the door suggests two related methods
for drawing an ellipse which are illustrated in the next pair of
diagrams. Both show the locus of D, defined as before as the
PF, + PF2= 2a
In the diagram below the origin is taken as midway between
the two foci, since this is obviously the centre of the ellipse.
midpoint of AB. They can be demonstrated with Cabri or
drawings can be made using simple apparatus as suggested
P is a general point on the ellipse with coordinates (xy) and
earlier. The first method uses a pair of hinged rods or strips,
with A moving along the x-axis. The second method is often
referred to as the 'slipping ladder' for obvious reasons. AC is
of constant length, with the point A moving along the x-axis
e is known as the eccentricity of the ellipse and takes values
between 0 and 1 corresponding to the two extremes of a circle
and the point C moving along the y-axis.
Mathematics
in
the coordinates of the foci are denoted as (- ae,0) and (ae,0).
and a straight line. Thus, the ellipse shown here has an
eccentricity which is about 1.2
School,
March
1999
7
y
However, y = + b when x = 0, where b is half the length of the
minor axis. It follows therefore that a2i (1 - e2) is equal to b2. On
P (X,y)
substituting b2, we have the familiar Cartesian equation for
an ellipse:
F,
F2
x
2
x
(-aeO) (ae.0)
2
y
--+ =1
a2 b2
Conclusion
We can now find expressions for the two lengths PF, and
PF, and use them to find the equation of the ellipse. Using
Pythagoras' Theorem the lengths are given by:
In many textbooks the ellipse is defined as the locus ofa point
which moves so that there is a constant ratio (the eccentricity)
between its distance from a fixed point (a focus) to its
distance from a fixed line (a directrix). This relationship can
be expressed as PF,= ePD, where D is a point on the directrix,
PF = (x + ae)2 + y2 and PF2 = (x- ae)2+ y2
Subtracting these two equations gives the simple result:
and the equation of the ellipse can be derived from it. However, this may seem a rather obscure starting point. Another
possibility is to define an ellipse as a section ofa cone. This is
intuitively appealing, but unfortunately it does not lead to a
particularly straightforward way of deriving the equation of
PF2 -PF 2 = 4aex
Since this is a difference of two squares, and we know that
PF, + PF,= 2a, it follows by dividing that:
the curve.
PF, -PF2 = 2ex
Solving PF, + PF2 = 2a and PF,- PF,2= 2ex as a pair of
simultaneous equations then gives expressions for the two
lengths PF, and PF,:
The stretched circle and the string and pins construction
which I have considered in this article provide interesting
alternative approaches which build on intuitive ideas of locus
and are rich in links and applications. There are, of course,
many other properties of the ellipse that merit further study,
but these will have to await another article! F
PF1 = a + ex and PF2 a - ex
Finally, substituting for PF, in PF' = (x + ae)2 + y2 gives:
(a + ex)2- (x + + Y2
References
Cabri G0ometre II 1996 Texas Instruments.
Cundy, H.M. and Rollett, A.P. 1951 Mathematical Models, Oxford University Press, Oxford. A reprinted version of this book is available from
Tarquin Publications.
a2 + 2aex + e2 2 = x2 + 2aex + a2e2 + y2
a2(l - e2) = x2(1 - e2) + y2
2
Dividing both sides by a(1 - e2) gives + Y
a 2 2 (1_ e2)
Author
Doug French, School of Education, University of Hull, Hull HU6 7RX.
Its wk/orse in /verse!.
Part I
by John Hersee
In my collection of old textbooks there are a number of prob-
lems set in verse. Here are two of them. Readers are invited to
try their hands at solving them. On page 39, answers are given
and the methods of solution are discussed. Note that spelling
2. A castle wall there was, whose height was found
To be a hundred feet from th' top to th' ground;
Against the wall a ladder stood upright,
Of the same length the castle was in height.
and punctuation are as in the originals. Although Imperial
units may be used, no detailed knowledge of the system of
A waggish youth did the ladder slide
units is required to solve the problems. Further problems will
appear in the next few issues of Mathematics in School.
Now I would know how far the top did fall,
1. As I was beating on the forest grounds,
Details of the old textbooks used will be given at the end of
the final instalment.
The distance that she started up before
Was fourscore sixteen rods just, and no more:
Now this I'd have you unto me declare,
Author
John Hersee, 76 Pembroke Road, Bristol BS8 3EG.
How far they ran before they caught the hare?
Mathematics
By pulling out the ladder from the wall.
References
Up starts a hare before my two greyhounds:
The dogs, being light of foot, did fairly run,
Unto her fifteen rods, just twenty-one.
8
(The bottom of it) ten feet from the side:
in
School,
March
1999