# Projections

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```					                  Projections of Ellipses and Circles
by
Gabrielle Renee Schrantz
(Supervised by Professor Ioannis M. Roussos)

Department of Mathematics, Hamline University
Spring 2004

Abstract: In this paper we combine geometry and calculus to study the projections of ellipses
and circles in space onto planes. We find that, apart from the trivial cases, these projections are ellipses and
in a special nontrivial case circles. We give their explicit parametric equations. The results are very useful
in geometry, in studying all sorts of elliptical motions (e.g., planetary motions) and in determining the
domains in the x-y plane of certain surfaces in space with boundary the given ellipses or circles.

1. Main work and results
We consider an ellipse with semi-axes a > b > 0 and with center the origin of an
x-y system of axes. This has implicit equation
x2 y2
   1
a 2 b2
A parameterization of this ellipse (in the positive direction) is
x  a cost 
 0  t  2 
y  b sin t 
(If a  b then we have a circle of radius r  a ( b) .)

1
Figure 1
The relation between the parameter-angle and the central angle  (as in figure 1) is
found by the relations:
d  a 2 cos 2 t  b 2 sin 2 t
x  d cos  a cos t
y  d sin   b sin t
So,
a                 b               b
cos       cost , sin   sin t , tan   tan t
d                 d               a
We observe that 0    t  2 .

Now suppose we have an ellipse with semi-axes a and b on a plane E  in space

and we want to determine its orthogonal projection onto a plane E  in space. Without

loss of generality, we consider E  to pass through the center of the ellipse. For this

work apart from the semi-axes a and b, we also need the angle 0              between the
2


two planes E  and E  and the angle 0           between the major axis of the ellipse
2

and the straight line of the intersection of E  and E  .




2


Figure 2

Since the projection lies on E  , we introduce coordinate x  axis to be the
straight line E   E  and y  axis the straight line on E  perpendicular to the x  axis


at the center of the ellipse, which is the origin (O) of the x  y system of axes in E  (see
figure 2). The projection of the point P, as in the figure 2, is let us say K= ( x , y ) . We now
have that d = length of OP, x  OA  d cos and y  AK  d sin  cos , with angle
  PAK . The central angle is      .
If we know the equations Ax  By  Cz  D of (E) and Ax  By  C z  D of
                                                                      

E  in some space coordinate system x-y-z with standard basis i , j , k , then we can
solve these two equations to find the straight line of their intersection. This is the x -axis

of the new x - y - z coordinate system, in which the x  y plane is E  . For instance, if
(E), the plane that contains the ellipse in the standard position as above, is the x-y plane,
that is z = 0, then D = 0 and the straight line E   E  has parametric equations:

x  (cos )t , y  (sin )t , z  0,     t  
The angle  satisfies
C
cos  
A 2  B  2  C  2

  
                                               
  

The standard basis i , j , k  of the x - y - z system is given in terms of i , j , k by
                      
i   cos i  sin  j
             
      A i  B j  C  k
k 
A 2  B 2  C  2
  
j   k   i

Going back to the problem of determining the projection of the ellipse, we have:

                          
x  d cos   dcos(   )
y  d cos sin   d cos sin(   )

So,
x  d(cos cos  sin sin)  acos cost  bsin sint

        Hence,
x  a 2 cos2   b 2 sin 2  cos(t   )

3
where  is the angle determined by

acos                               bsin 
cos                                , sin 
a2 cos2   b2 sin2                a2 cos2   b2 sin2 

Similarly,
y  cos (b cos sin t  a sin  cost )  cos a 2 sin 2   b 2 cos2  sin(t   )

where  is the angle determined by

bcos                               asin
cos                            , sin  
a sin   b cos 
2    2       2   2
a sin   b2 cos2 
2   2

We observe that
b             a
tan  tan, tan   tan
a             b

Now we are in the position to conclude the following results:

Case of a circle of radius r  a ( b)

In the case of a circle, we can take       0 , since a  b and any diameter may be
considered as major axis. So, we get
x  r cost 
 0  t  2
y  r cos sin t 
Hence:

1.) If   0,                                                                                 
, this case is not trivial. The projection of such a circle in E  onto E  is
2
an ellipse in the standard position with semi-axes:
a = r and b = r cosω


2.) If   0 , then E   E  . This case is trivial and the circle projects onto itself.

3.) If     , then E   E  . This case is trivial, because y  0 for all the values of t
2
and so the circle projects onto the straight segment r  x  r of the x  axis.

Case of ellipse (a  b) .

4

1.) If   0 , then E   E  . We can take   0 . This case is trivial and the ellipse pro-
jects onto itself.

2.) If     , then E   E  . This case is trivial, because y  0 for all the values of t
2
and so the ellipse projects onto the straight segment of the x  axis

 a 2 cos2   b2 sin2   x  a2 cos2   b2 sin2 

3.) Suppose 0               . This case is non-trivial and we have the following three intere-
2
sting sub-cases:
3a.) If   0 , then the projection on the ( E plane is an ellipse in the standard position,
)
with respect to the x  y system of axes, with major axis the x  axis . In fact we have
      0 and
x  a cost 
 0  t  2
y  b cos sin t 

Hence, the semi-axes of the projection are a > bcos .

3b.) If         , then the projection on the ( E plane is an ellipse in the standard position,
)
2

with respect to the x  y system of axes. In fact we have                      and
2
x  b sin t    
 0  t  2
y  a cos ( cost )

b
Hence, this ellipse has semi-axes b and a cosω. Consequently, if              cos , then b is the
a
b
major semi-axis and if  cos , then b is the minor semi-axis.
a

b
               Finally, if cos  , then the projection is a circle. This is the only nontrivial
a

case in which the respected projection is a circle (that is, when              and b = a cosω).
2

3c.) If 0           , then the projection on the ( E plane is an ellipse parameterized by
)
2
x  a 2 cos 2   b 2 sin 2  cos(t   )   
 0  t  2
                          y  cos  a 2 sin 2   b 2 cos 2  sin(t   )


5
This ellipse has center the origin of the x  y system of axes, but it is not in the standard
position, since in this case    and so we have a phase-difference between the x and
the y coordinates of any point P  ( x , y ) of this projection. This ellipse is tilted.

To understand this sub-case fully we must first examine in some detail the
equations

x  A cos(t   )
 0  t  2
y  B sin t 

where 0         is the phase-shift and A > 0, B > 0 are constants.
2
If   0 , then we get the ellipse
x  A cost 
 0  t  2
y  B sin t 
in the standard position.

If          , then we get
2
x   A sin t 
 0  t  2
y  B sin t 
This is a straight segment joining the points (A, -B) and (-A, B) of the x-y plane with
B
slope      , traversed twice starting from the origin (0, 0).
A

If 0    , then we eliminate the parameter t between the two equations and
2
we reach the equation
                     1 2 2 sin             1
2
x            xy  2 y 2  cos2 
A           AB         B

Note that this equation, by the theory of the general second-degree equation in two

variables, represents an ellipse. We observe that it is also valid for   0 or     . To
2
obtain the standard position of the ellipse with respect to a rotated system of axes we

need the angle of rotation  (mod ) to be given by the relation
2

6
2 sin 
tan 2    AB  2 AB sin 
1      1 B 2  A2
 2
A2 B

         
Notice that if A = B > 0 then  =     (mod ) regardless of   0 .
4        2
Now the ellipse found in (3c.) can be written as

x  A cos(t   )
 -   t  2  
y  B sin t 
with

t  t 

  

A  a 2 cos2   b 2 sin 2 

B  cos a 2 sin 2   b 2 cos2 


Notice that for 0          and a  b  0 , by using the expressions for the tangents of
2
 and  , stated earlier, we find that

a 2  b 2 tan 
0  tan  tan(   )                    
ab 1  tan 2 

and so


0  
2

Now, using the expressions for sine and cosine of  and  , stated earlier, we find

that the angle  (mod ) of the rotation of the axes in the ( E plane is found from the
)
2
equation

7
2(a 2  b 2 ) cos sin  cos
tan 2 
cos2  (a 2 sin 2   b 2 cos2  )  (a 2 cos2   b 2 sin 2  )

So,  depends on the parameters a, b,  and  . This formula also agrees with the results
in the other cases.

8
1 2 2 sin      1
A note about the lengths of the semi-axes of                       2
x     xy  2 y 2  cos2 
A       AB     B

We compute the two positive eigenvalues 0  1  2 of the 2x2 symmetric and positive
definite matrix of the coefficients of this second degree equation

 1                 sin  
 A2                 AB 
 sin                1 
                         
 AB                 B2 

After the rotation, in the new u-v coordinate system, the equation transforms to

1u 2   2 v 2  cos 2 

or to the standard form

u2                      v2
2
                   2
1
 cos               cos      
                              
                            
    1                  2      


Then, with 0          , the lengths of the semi-axes are
2

cos
Major semi - axis 
1

cos
Minor semi - axis 
2

The rotation matrix that achieves the above transformation, from the x-y coordinates to u-
v coordinates, is precisely the 2x2 matrix with columns the unit eigenvectors correspon-
ding to the eigenvalues 1 and 2 of the initial matrix.

9
2. Parameterization of an Ellipse (or Circle) in Space
To parameterize an ellipse in space we need to know:

1) Its center C  ( q1, , q 2 , q 3 )
2) The plane on which it lies ax  by  cz  d (with normal N  ai  bj  ck )

3) The direction of the major axis u . We consider u  1 . (Also, u  N  0)
4) The lengths of the major and minor semi-axes a and b .

The direction of the minor axis is found by:
w
v
w
where
  
wuN



        

Figure 3

10
Then any point P  ( x, y, z ) of the ellipse satisfies
           
( x, y, z )  (q1 , q 2 ,q3 )  a cos t u  b sin t v ,   0  t  2

Writing this as x  x(t ), y  y(t ), z  z(t ) we have a parameterization of the ellipse.
If we have a circle of radius r ( a  b) , then we simply find two orthonormal

solutions of ax  by  cz  0, u and v , and since any diameter of the circle is major axis
we have:
         
( x, y, z )  (q1 ,q 2 ,q3 )  r (cos t u  sin t v ) , 0  t  2

3. Derivation of Equations by Vector Calculus
We can now re-derive the previous equations using purely vector calculus, that is,
without much need of the geometry of figures 1 and 2. However, some geometrical
suggestions are still present.
Let again x  axis be the intersection of (E) and ( E and y  axis the line
)
perpendicular to it at the center of the ellipse (or circle) and lying in ( E . Here we write
)
x, y instead of x, y . So, we have in  the usual basis i , j , k  i  j . Since the ellipse has
3

center the origin, if u, v are its axial directions, its parameterization is given by
                     
r  a cost u  b sin t v , 0  t  2
or
                                                              
xi  yj  zk  a cost (u1i  u 2 j  u3 k )  b sin t (v1i  v2 j  v3 k )
  
where u1 , u 2 , u 3 and v1 , v 2 , v3 are the components of u and v respectively. So
component-wise we have
                                              x  au1 cost  bv1 sin t 

y  au 2 cost  bv2 sin t  0  t  2
z  au3 cost  bv3 sin t 

Therefore, the projection of the ellipse onto the x  y plane is the curve with
parameterization the first two equations

x  au1 cost  bv1 sin t 
         0  t  2
y  au 2 cost  bv2 sin t 

If these equations for x and y are transformed to the equations found earlier then the
remaining work is exactly the same as done before.
11
         

We need to replace the u , v1, u2, v2 . We have that u, v , i are in the plane (E) and
1
                             
let angle of (u , i )   . So angle of (v , i )    . Then
2
u1  i  u  cos

v1  i  v  cos (   )  sin
2

Hence

x  a cos cost  b sin  sin t

which is the same equation as before for x .

Next we have that
               
u 2  j  u and v 2  j  v
To evaluate these we need the fact that u  v  n , the unit normal to (E) , and so
n  k  cos , where ω is the angle between (E) and ( E .
)
We now have the following straightforward computations:
                                  
i  u  sin  (n )   sin  n                   (notice the – sign here)
  
(i  u )  k   sin  cos                        (i  j  k )
   
[( j  k )  u ]  k   sin  cos                (by the familiar identity of vector calculus)
      
[(u  k )  j  (u  j )  k ]  k   sin  cos 
0  u  j  sin  cos
finally u 2   sin  cos
Also
                                        
i  v  sin(   ) n  cos n
2
(i v )  k  cos cos
   
[( j  k )  v ]  k  cos  cos 
      
[(v  k )  j  (v  j )  k ]  k  cos cos 
 
0  v  j  cos cos 
finally v2  cos cos
So we get

y  cos (b cos sin t  a sin  cost )

which is the same equation as before for y (done!).

12

An Interesting Example from the Classical Euclidean Geometry
The Steiner’s Ellipse of a Triangle
It is a fairly straightforward fact that every triangle ABC is the orthogonal proje-
ction of a unique equilateral triangle ABC , up to the parallel translations perpendicular
to the plane ABC (prove this!). Then, the circumscribed circle of ABC projects onto
the plane of ABC as an ellipse, in general, that passes through the points A, B and C. This
ellipse is called the Steiner’s Ellipse of the triangle ABC, after the German Mathemati-
cian J. Steiner (1796-1863), who introduced it first. This ellipse was studied by a good
many and famous mathematicians in the years 1880-1900, for the properties that it posse-
sses. Here we cite two of them:

1) Steiner himself proved that this ellipse besides containing the points A, B, C of
the circumscribed circle of the triangle ABC, it also contains another fourth point
of this circle, called Steiner’s point. This point and its diametrically opposite one
(Tarry’s point) have many properties that have been studied extensively.

Area of St einer's El lipse   4
2)                                                        
Area of Co rresponding Triangle 3 3

(fixed ratio, independent of the initial triangle).

For some details and a lot references on this matter, see the authoritative work:
F. M. G., Exercises De Géométrie, Librairie Générale, Paris, Septiéme Edition,
pp. 1112-1114.

Miss Gabrielle Schrantz (MB 1463)
Department of Mathematics
1536 Hewitt Ave.
Hamline University
Saint Paul, MN, 55104

E-mails:
gschrantz01@gw.hamline.edu
iroussos@gw.hamline.edu

13

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