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E'A' § E'C' = [ EA § EC ( 1 ’ r )2 + EA § EC v § OE r ( 1 ’ r ) ( v § OF ) ’1 ]

[1’ r v § AF (v § OF ) ’1 ] ’1 [1’ r v § CF ( v § OF ) ’1 ] ’1 [1’ r v § EF ( v § OF ) ’1 ] ’2

Extracting common factor:

E'A' § E'C' = EA § EC [ ( 1 ’ r )2 + v § OE r ( 1 ’ r ) ( v § OF ) ’1 ]

[1’ r v § AF (v § OF ) ’1 ] ’1 [1’ r v § CF ( v § OF ) ’1 ] ’1 [1’ r v § EF ( v § OF ) ’1 ] ’2

In the projective cross ratio, all the factors are simplified except the first outer product:

E'A' § E'C' E'B' § E'D' EA § EC EB § ED
=
E'A' § E'D' E'B' § E'C' EA § ED EB § EC

This proves that the projective cross ratio of any four points A, B, C and D with respect to
a centre of projection E is equal to the projective cross ratio of the homologous points A',
B', C' and D' with respect to the homologous centre of projection E'. It means that the
homology is a special kind of projectivity where a line is preserved, the axis of homology.
When the centre of projection is a point H on the homology axis, H and H' are coincident
and we find in the former equality the initial condition of the homology again:

H = H' { H, A' B' C' D' } = { H, A B C D }

The following chapter is devoted to the conics and the Chasles™ theorem, which
states that the locus of the points from where the projective cross ratio of any four points
is constant is a conic passing through these four points. Since the homology preserves the
projective cross ratio, it transform conics into conics. Hence, a conic may be drawn as the
homologous curve of a circle. Depending on the position of this circle there are three
cases:
1) The circle does not cut the limit line L (whose homologous is the line at
infinity). In this case the homologous curve is an ellipse, because very next points on the
circle have also very next homologous points and therefore the homologous curve must be
closed.
2) The circle touches the limit line L. Then its homologous curve is the parabola,
because the homologous of the contact point is a point at infinity. The symmetry axis of
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 115

the parabola has the direction of the line passing through the centre of homology and the
contact point.
3) The circle cuts the limit line L. Then the homologous curve is the hyperbola
because the homologous points of both intersections are two points at infinity. The lines
passing through the centre of homology and both intersection points have the direction of
the asymptotes.


Exercises

10.1 Prove the Ptolemy™s theorem: For a quadrilateral inscribed in a circle, the product of
the lengths of both diagonals is equal to the sum of the products of the lengths of the pairs
of opposite sides.

10.2 Find that the point D forming a harmonic range with A, B and C is given by the
equation:

D = A + AB ( 1 ’ 2 AC ’1 BC ) ’1

10.3 Show that the homography is a directly conformal transformation, that is, it preserves
angles and their orientations.

10.4 Prove that if A', B', C' and D' are the homologous of A, B, C and D, and H is a point
on the homology axis, then the equality { H, A' B' C' D' } = { H, A B C D } holds. That is,
show directly that:

HA' § HC' HB' § HD' HA § HC HB § HD
=
HA' § HD' HB' § HC' HA § HD HB § HC

10.5 A special conformal transformation of centre O and translation v is defined as the
geometric operation which, given any point P, consists in the inversion of the vector OP,
the addition of the translation v and the inversion of the resulting vector again.

OP' ’1 = ( OP ’1 + v ) ’1

Prove the following properties of this transformation:
a) It is an additive operation with respect to the translations: the result of
applying firstly a transformation with centre O and translation v, and later a
transformation with the same centre and translation w is identical to the result
of applying a unique transformation with translation v + w.
b) It preserves the complex cross ratio and thus it is a special case of
homography.

10.6 Prove that if a homography keeps invariant three or more points on the plane, then it
is the identity.
RAMON GONZALEZ CALVET
116

10.7 An antigraphy is defined as that transformation which conjugates the complex cross
ratio of any four points:

( A B C D ) = ( A' B' C' D' )*

Then an antigraphy is completely specified by giving the images A', B', C' of any three
points A, B, C. Prove that:
a) The antigraphy is an opposite conformal transformation, that is, it changes the
orientation of the angles but preserving their absolute values.
b) The composition of two antigraphies is a homography.
c) A product of three inversions is an antigraphy.
d) If an antigraphy has three invariant points, then it is a circular inversion.

10.8 Let A, B, C and A', B', C' be two sets of independent points. A projectivity has
been defined as the mapping:


D=aA+bB+cC D' = a' A' + b' B' + c' C'

where the coefficients a', b', c' are homogeneous linear functions of a, b, c. Prove that
collinear points are mapped to collinear points.

10.9 Prove the following theorem: let the sides of a hexagon ABCDEF pass alternatively
through the points P and Q. Then the lines AD, BE and CF joining opposite vertices meet
in a unique point. Hint: draw the hexagon in the dual plane.
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 117

11. CONICS

Conic sections

The conic sections (or conics) are the intersections of a conic surface with any
transversal plane (figure 11.1). The proper conics are the curves obtained with a plane not
passing through the vertex of the cone. If the plane contains the vertex we have improper
conics, which reduce to a pair of straight lines or a point. In general, I shall regard only
proper conics, taking into account that the other case can be usually obtained as a limit
case.
Let us consider the two spheres inscribed in the cone which are tangent to the




Figure 11.1




plane of the conic. The spheres touch this plane at the points F and F', the foci of the conic
section. On the other hand, both spheres touch the cone surface at two tangency circles
lying in planes perpendicular to the cone axis. The directrices of the conic are defined as
the intersections of these planes with the plane of the conic.
Let P be a point on the conic. Since it lies on the plane of the conic, the segment
going from P to the focus F is tangent to the sphere. Since it belongs to the cone surface,
the generatrix passing through P is also tangent to the sphere. PF and PS are tangent to the
same sphere so that their lengths are equal: ¦PF¦=¦PS¦. Also PF' and PS' are tangent to
the other sphere so that their lengths are also equal: ¦PF'¦=¦PS'¦. On the other hand the
segment SS' of any directrix has constant length, what implies that the addition of
distances from any point P on the conic to both foci is constant:
RAMON GONZALEZ CALVET
118

PF + PF' = PS + PS' = SS' = constant

This discussion slightly changes for a plane which intercepts the upper and lower cones.
In this case, the distances from P to the upper sphere and to the upper branch of the curve
must be considered negative1. If the plane is parallel to a generatrix of the cone these
distances become infinite. This unified point of view for all conics using negative or
infinite distances when needed will be the guide of this chapter. I shall not separate
equations but only specify cases for distinct conics.
A lateral view of figure 11.1 is given in the figure 11.2. In this figure the point P
has been drawn in both extremes of the conic (and S and S' also twice), although we must
consider any point on the conic section. As indicated previously ¦PF¦=¦PS¦. But ¦PS¦
is proportional to the distance from P to the upper plane containing S and this distance is
also proportional to the distance from P to the directrix r (figure 11.2):

sin±
d (P , r )
PF = PS =
sinδ

The quotient of the sines of both angles is called the eccentricity e of the conic:

sin±
e= >0 Figure 11.2
sinδ


Then a conic can be also defined as
the geometric locus of the points P
whose distance from the focus F is
proportional to the distance from the
directrix r (figure 11.3).

FP = e d (P, r )

¦FP¦ is called the focal radius. Let T
be the point of the directrix closest to
the focus (figure 11.3). Then the main
axis of symmetry of the conic is the line
which is perpendicular to the directrix and
passes through the focus. Under a
reflection with respect the main axis of
symmetry, the conic is preserved. The
oriented distance from any point P to the
directrix r is expressed, using the vector
FT, as:
Figure 11.3


1
To consider sensed distances is not a trouble but an advantage. For example, the distance from
a point to a line can be also taken as an oriented distance, whose sign indicates the half-plane
where the point lies.
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 119

FT · PT
d (P , r ) =
FT

Defined in this way, the distance from P to r is positive when P is placed on the half-plane
that contains the focus of both in which the directrix divides the plane, and negative when
P lies on the other half-plane. The equation of the conic becomes:

FT · PT
FP = e d (P, r ) = e
FT
FP FT = e FT · PT = e FT · (FT ’ FP ) = e (FT 2 ’ FT · FP )

Let us denote by β the angle between the main axis FT and the focal vector FP:

FP FT = e (FT 2 ’ FT FP cos β )

FP FT (1 + e cos β ) = e FT 2

From here one obtains the polar equation of a conic:

FT
FP =
1 + e cos β

According to the value of the eccentricity there are three types of proper conics:
1) For 0 < e < 1, the inclination ± of the plane of the conic is lower than the
inclination δ of the cone generatrix and the denominator is always positive.
Then the focal radius is always positive and the points P form a closed curve
on the focal half-plane, an ellipse.
2) For e = 1, ± = δ , the plane of the conic is parallel to a cone generatrix. Since
the denominator vanishes for β = π, the points P form an open curve called
parabola. Except for this value of β, the radius is always positive and the
curve lies on the focal half-plane.
3) For e > 1, ± >δ , the plane of the conic intercepts both upper and lower cones.
The denominator vanishes twice so that it is an open curve with two branches
called hyperbola (figure
11.4). The lowest positive
Figure 11.4 angle that makes the
denominator zero is the
asymptote angle βa :

« 1
β a = arccos¬ ’ ·
 e

The maximum eccentricity of the
hyperbolas as cone sections is obtained
with ± =π/2:
RAMON GONZALEZ CALVET
120

1
1< e ¤
sinδ

The focal radius is positive (P lies on the focal branch) for the ranges of β :

0 < β < βa 2π “ βa < β < 2π
and

and negative (P lies on the non focal branch) for the range:

βa < β < 2π “ βa

A negative focal radius means that we must take the sense opposite that determined by the
angle (figure 11.4). For example, a radius “2 with an angle 7π/6 is equivalent to a radius 2
with an angle of π/6. So the vertex T' of the non focal branch has an angle π (not zero as it
would appear) and a negative focal radius.


Two foci and two directrices Figure 11.5

There are usually two spheres
touching the cone surface and the conic
plane, and hence there are also two foci. For
the case of the ellipse, both tangent spheres
are located in the same cone. For the case of
the hyperbola, each sphere is placed in each
cone so that each focus is placed next to
each branch. In the case of the parabola,
one sphere and the corresponding focus is
placed at the infinity. For the improper
Figure 11.6
conics, the tangent spheres have null radius
and the foci are coincident with the vertex of
the cone, which is the crossing point of both
lines.
As the figure 11.1 shows for a proper
conic, both directrices r and r' are parallel, and
both foci F and F' are located on the axis of
symmetry of the conic. Above we have already
seen that the addition of oriented distances
from both foci to any point P is constant. Let
us see the relation with the distance between
the directrices. If P is any point on the conic
then:

FP + F'P = e d (P, r ) + e d (P, r' ) = e d (r, r' )

The sum of the oriented distances from any point P to any two parallel lines (the
directrices in our case) is constant2. Then the addition of both focal radii of any point P

2
Note that this statement is only right for oriented distances but not for positive distances.
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 121

on the conic is the product of the eccentricity multiplied by the oriented distance between
both directrices. P can be any point, for example Q, the extreme of the conic:
FP + F'P = FQ + F'Q = F'Q' + F'Q = QQ'

whence it follows that the addition of both focal radii is QQ' , the major diameter.
Specifying:
1) For the case of the ellipse (figure 11.5), both focal radii are positive and also
the major diameter.
2) For the case of the parabola, one directrix is the line at the infinity and one
focus is a point at the infinity on the main axis of symmetry. Hence the major
diameter has an infinite value.
3) For the case of the hyperbola (figure 11.6), the focal radius of a point on the
non focal branch is negative, its absolute value being higher than the other
focal radius, which is positive, yielding a negative major diameter.


Vectorial equation

The polar equation of a conic may be written having as parameter the focal
distance ¦FQ¦ instead of the distance from the focus to the directrix ¦FT¦:

1+ e
FP = FQ
1 + e cos β

Let Q' be the vertex of the conic closest to the focus F'. The distance from the focus F to
Q' is found for β = π:

1+ e
FQ' = FQ
1’ e

For the ellipse ¦FQ'¦ is a positive distance; for the hyperbola it is a negative distance and
for the parabola Q' is at infinity. Then the distance between both foci is:

2e
FF' = FQ' ’ F'Q' = FQ' ’ FQ = FQ
1’ e

On the other hand the major diameter is:

2 FQ
QQ' = FQ + FQ' =
1’ e

By dividing both equations, an alternative definition of the eccentricity is obtained:

FF'
e=
QQ'
RAMON GONZALEZ CALVET
122

The eccentricity is the ratio of the distance between both foci divided by the major
diameter. For the case of ellipse, both distances are positive. For the case of the hyperbola,
both distances are negative, so the eccentricity is always positive. When e = 0 both foci
are coincident in the centre of a circle. Note that the circle is obtained when we cut the
cone with a horizontal plane. In this case, the directrices are the line at the infinity.
The vectorial equation of a conic is obtained from the polar equation and contains
the radius vector FP. Since FP forms with FQ an angle β (figure 11.3), FP is obtained
from the unitary vector of FQ via multiplication by the exponential with argument β and
by the modulus of FP yielding:

1+ e
FQ (cos β + e12 sinβ )
FP =
1 + e cos β

On the other hand, from the directrix property, one easily finds the following
equation for a conic:

FP2 FT2 = e2 ( FT2 ’ FT · FP )2

F, T and e are parameters of the conic, and P = (x, y) is the mobile point. Therefore from
this equation we will also obtain a Cartesian equation of second degree. For example, let
us calculate the Cartesian equation of an ellipse with eccentricity ½ and focus at the point
(3, 4) and vertex at (4,5):

F = ( 3, 4 ) Q = ( 4, 5 ) P = ( x, y )
e = 1/2

1+ e
FT = FQ = 3 e1 + 3 e 2 T = F + FT = ( 6, 7 )
e

FP = ( x ’ 3 ) e1 + ( y ’ 4 ) e2

Then the equation of the conic is:

1
[ ( x ’ 3 )2 + ( y ’ 4 )2 ] 18 = [ 18 ’ ( 3 (x ’ 3) + 3 ( y ’ 4 ) ) ]2
4

After simplification:

7 x2 ’ 2 x y + 7 y2 ’ 22 x ’ 38 y + 31 = 0


The Chasles™ theorem

According to this theorem3, the
projective cross ratio of any four given
points A, B, C and D on a conic with
respect to a point X also on this conic is
Figure 11.7
constant independently of the choice of
3
Michel Chasles, Trait© des sections coniques, Gauthier-Villars, Paris, 1865, p. 3.
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 123

the point X (figure 11.7):
{ X, A B C D } = { X', A B C D }

Figure 11.8
To prove this theorem, let us take into
account that the points A, B, C, D and X must
fulfil the vectorial equation of the conic. Let
us also suppose, without loss of generality,
the main axis of symmetry having the
direction e1 (this supposition simplifies the
calculations):

FQ = ¦FQ¦ e1

Now on ±, β, γ, δ and χ will be the angles
which the focal radii FA, FB, FC, FD and
FX form with the main axis with direction
vector FQ (figure 11.8). Then:

® e cos ± + e 2 sin± e1 cos χ + e 2 sinχ 
XA = FA ’ FX = FQ (1 + e )  1 ’ 
1 + e cos ± 1 + e cos χ
° »

Introducing a common denominator, we find:

(1 + e ) [e1 (cos ± ’ cos χ ) + e 2 (sin± ’ sinχ + e sin± cos χ ’ e cos ± sinχ )]
XA = FQ
(1 + e cos ± ) (1 + e cos χ )

From XA and the analogous expression for XC, and after simplification we obtain:

e (1 + e ) (sinγ cos ± ’ sin± cos γ + sinχ cos γ ’ sinγ cos χ + sin± cos χ ’ sinχ cos ± )
2

XA § XC = FQ 122

(1 + e cos ± )(1 + e cos γ )(1 + e cos χ )

(1 + e )2 [ sin(γ ’ ± ) + sin (χ ’ γ ) + sin (± - χ ) ]
= FQ 2
e12
(1 + e cos ± )(1 + e cos γ )(1 + e cos χ )

Using the trigonometric identities of the half-angles, the sum is converted into a product
of sines (exercise 6.2):
( 1 + e )2 ® sin« γ ’ ±  sin« χ ’ γ  sin« ± ’ χ  
 ¬ 2 · ¬ 2 · ¬ 2 ·
°   »
XA § XC = ’ 4 FQ 2
e12
( 1 + e cos ± ) ( 1 + e cos γ ) ( 1 + e cos χ )

In the same way the other outer products are obtained. The projective cross ratio is their
quotient, where the factors containing the eccentricity or the angle χ are simplified:
RAMON GONZALEZ CALVET
124

γ ’± δ ’β AFC BFD
sin sin sin sin
XA § XC XB § XD 2 2= 2 2
{X , A B C D } = =
δ ’± γ ’β
XA § XD XB § XC AFD BFC
sin sin sin sin
2 2 2 2
since γ ’± is the angle AFC, etc. Therefore, the projective cross ratio of four points A, B,
C and D on a conic is equal to the quotient of the sines of the focal half-angles, which do
not depend on X, but only on the positions of A, B, C and D, fact which proves the
Chasles™ theorem. This statement is trivial for the case of the circumference, because the
inscribed angles are the half of the central angles. However the inscribed angles on a conic
vary with the position of the point X and they differ from the half focal angles. In spite of
this, it is a notable result that the quotient of the sines of the inscribed angles (projective
cross ratio) is equal to the quotient of the half focal angles. For the case of the hyperbola I
remind you that the focal radius of a point on the non focal branch is oriented with the
opposite sense and the focal angle is measured with respect this orientation.


Tangent and perpendicular to a conic

The vectorial equation of a conic with the
Figure 11.9
major diameter oriented in the direction e1 (figure
11.9) is:

(1 + e ) FQ
( e1 cos ± + e 2 sin± )
FP =
1 + e cos±

The derivation with respect the angle ± gives:

( 1 + e ) FQ
d FP
[’ e sin± + e2 ( e + cos ± ) ]
=
( 1 + e cos ± )2 1


This derivative has the direction of the line tangent to the conic at the point P, its unitary
vector t being:

’ e1 sin± + e 2 ( e + cos ± )
t=
1 + e 2 + 2 e cos ±

The unitary normal vector n is orthogonal to the tangent vector:

e1 ( e + cos ± ) + e 2 sin±
n=
1 + e 2 + 2 e cos ±

In the same way, t and n are obtained as functions of the angle β (figure 11.9) from the
vectorial equation for the focus F':
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 125

(1 ’ e ) FQ
( e1 cos β + e 2 sinβ )
F'P =
1 ’ e cosβ

’ e1 sinβ + e 2 (’ e + cos β )
t=
1 + e 2 ’ 2 e cos β Figure 11.10

e1 ( ’ e + cos β ) + e 2 sinβ
n=
1 + e 2 ’ 2 e cos β

Let us see now that the normal
vector has the direction of the bisector of
the angle between both focal radii (figure
11.10). In order to prove this statement, let
us consider an infinitesimal displacement
of a point P on the conic (figure 11.11).
The sum of the focal radii of any conic is constant, whence it follows that the addition of
their differentials are null:

¦FP¦ + ¦F'P¦ = constant ’ 0 = d¦FP¦ + d¦F'P¦

The differential of the focal
4
Figure 11.11
radius is obtained by differentiating the
square of the focal vector:

d FP2 = 2 FP · d FP = 2 ¦FP¦ d¦FP¦

d FP · FP
d FP = ’
FP

Summing the differentials of both
moduli we obtain:

« FP F'P 
d FP · FP d F'P · F'P
= ’ d FP · ¬ ·
0 = d FP + d F'P = ’ ’ +
¬ FP F'P ·
FP F'P  

since the differentials of the focal vectors are equal: dP = d FP = d F'P. In conclusion the
differential vector (or the tangent vector) is orthogonal to the bisector of both focal
vectors. The figure 11.11 clearly shows that both right triangles are opposite because they
share the same hypotenuse d FP, and the legs d¦FP¦ and d¦F'P¦ have the same length.
The reflection axis of both triangles, which is the shorter diagonal of the rhombus, is the
bisector line of both focal vectors.
In the case of the parabola (figure 11.12), the focus F' is located at the infinity and
F'P is parallel to the main axis of symmetry. In the case of the hyperbola (figure 11.13)
we must take into account that the radius of a point on the non focal branch is negative, so

4
Note that it differs from the modulus of the differential of the focal vector: d¦FP¦≠¦d FP¦.
RAMON GONZALEZ CALVET
126

it has the sense from P to the focus F'. The normal direction is the bisector of both
oriented focal vectors.
These geometric features are the widely known optic reflection properties of the
conics: parallel beams reflected by a parabola are concurrent at the focus. Also the beams
emitted by a focus of an ellipse and reflected in its perimeter are concurrent in the other
focus.



Figure 11.12 Figure 11.13




Central equations for the ellipse and hyperbola

We search a polar equation to describe a conic using its centre as the origin of the
radius. Above we have seen that the eccentricity is the ratio of the distance between both
foci and the major diameter:

FF'
e=
QQ'

If O is the centre of the conic then it is the
midpoint of both foci:

F + F'
O=
2

OF FQ
e= 1’ e = Figure 11.14
OQ OQ

For the case of the ellipse all the quantities are positive. For the case of the
hyperbola both ¦OF¦ and ¦OQ¦ are negative with ¦FQ¦ positive. Let the angle QOP be
γ (figure 11.14). Then:

¦OP¦ sin γ = ¦FP¦ sin ±

¦OP¦ cos γ = ¦OF¦ + ¦FP¦ cos ±
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 127

For the ellipse all the moduli are positive whereas for the hyperbola ¦OP¦ and
¦OF¦ are always negative (the centre is placed between both branches). For the parabola
they become infinite. By substitution of the polar equation into these equalities we have:

(1 ’ e )sin±
(1 + e ) sin± 2
OP sinγ = FQ = OQ
1 + e cos ± 1 + e cos ±

(1 ’ e 2 )cos ±  = OQ e + cos±
FQ (1 + e ) cos± ®
OP cosγ = OF + = OQ e + 
1 + e cos ± 1 + e cos ± » 1 + e cos ±
°

Summing the squares of both former equalities we find:

« sin 2 γ  OQ
+ cos 2 γ · = OQ 2
OP ¬ OP =
2

¬1 ’ e2 ·
sin 2 γ
 
+ cos 2 γ
1’ e 2




The direction of OP with respect to OQ is given by the angle γ and we may add it
multiplying by the unitary complex with argument γ to obtain the central vectorial
equation:

OQ (cos γ + e12 sinγ )
OP =
sin 2 γ
+ cos 2 γ
1’ e 2




Let R be the point P for γ =π/2. Then OR lies on the secondary axis of symmetry
of the conic, which is perpendicular to OQ, the main axis of symmetry. ¦OR¦is the minor
half-axis. The eccentricity relates both:

OR2 = ( 1 ’ e2 ) OQ2

OR = 1 ’ e 2 OQ


OR = 1 ’ e 2 OQ e12

Using the minor half-axis, the equation of the ellipse becomes:

« sinγ 
1
¬ OQ cos γ + OR ·
OP = ¬ 2·
sin 2 γ 1’ e 

+ cos 2 γ
1’ e 2




Hence it follows:

OP 2 OP 2
sin γ + cos 2 γ = 1
2
2 2
OR OQ
RAMON GONZALEZ CALVET
128



Taking OP and OQ as the coordinates axis and O the origin of coordinates, we have the
canonical equation of a conic:

x = ¦OP¦ cos γ y = ¦OP¦ sin γ

x2 y2
+ =1
OQ 2 OR 2

This equation of a conic is specific of this coordinate system and has a limited
usefulness. If another system of coordinates is used (the general case) the equation is
always of second degree but not so beautiful. For the ellipse both half-axis are positive
real numbers. However for the hyperbola, the minor half-axis ¦OQ¦ is imaginary and
OQ2<0 converting the sum of the squares in the former equation in a difference of real
positive quantities.


Diameters and Apollonius™ theorem

If e<1 (ellipse) we may introduce the angle θ in the following way:

Figure 11.15
cos γ
cos θ =
sin 2 γ
+ cos 2 γ
1’ e 2




Then the central equation of an ellipse
becomes5:

OP = OQ cos θ + OR sin θ

The angle θ has a direct geometric
interpretation if we look at a section of a
cylinder (figure 11.15):

cos γ cos φ
cos θ =
sin 2 γ + cos 2 γ cos 2 φ

From where it follows the relationship between the inclination of the cylindrical section
and the eccentricity:

cos φ = 1 ’ e 2

5
G. Peano (Gli elementi di calcolo geometrico [1891] in Opere Scelte, vol. III, Edizioni Cremone,
[Roma, 1959], p.59) gives this central equation for the ellipse as function of the angle θ. He also
used the parametric equations of the parabola, hyperbola, cycloid, epicycloid and the spiral of
Archimedes.
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 129



Then θ is an angle between axial planes of the cylinder.
The equation of the ellipse may be written using another pair of axis turned by a
cylindrical angle χ (figure 11.16):

OQ' = OQ cos χ + OR sinχ OR' = ’OQ sin χ + OR sinχ

Each pair of OQ' and OR' obtained in this way are conjugate central radii and
twice them are conjugate diameters. They are intersections of the plane of the ellipse with
two axial planes of the cylinder that form
a right angle. From the definition it is
obvious that:

OQ' 2 + OR' 2 = OQ2 + OR2

The inverse relation between the
conjugate radii are:

OQ = OQ' cos χ ’ OR' sin χ

Figure 11.16
OR = OQ' sin χ + OR' cos χ

Then, the central radius of any point P is:

OP = ( OQ' cos χ ’ OR' sin χ ) cos θ + ( OQ' sin χ + OR' cos χ ) sin θ

= OQ' cos( θ ’χ ) + OR' sin( θ ’χ )

Changing the sign of the cylindrical angle we obtain another point P' on the ellipse owing
to the parity of trigonometric functions:

OP' = OQ' cos(θ ’χ ) ’ OR' sin(θ ’χ ) = OQ' cos(χ ’θ ) + OR' sin(χ ’θ )

Then the chord PP' is parallel to the radius OR' (figure 11.16):

PP' = OP' ’ OP = ’2 OR' sin( θ ’ χ )

Now, let us mention the Apollonius theorem: a diameter of a conic is formed by
all the midpoints of the chords parallel to its conjugate diameter. To prove this statement,
see that the central radius of the midpoint has the direction OQ':

OP + OP'
= OQ' cos(θ ’ χ )
2

The properties of the conjugate diameters are also applicable to the hyperbola, but
taking into account that OR2 = (1 ’ e2) OQ2 < 0, that is, the minor half-axis ¦OR¦ is an
imaginary number. In this case OR has the same direction than OQ:
RAMON GONZALEZ CALVET
130


OR = 1 ’ e 2 OQ e12 = e 2 ’ 1 OQ

And also the angle θ is an imaginary number, and the trigonometric functions are turned
into hyperbolic functions of the real argument ψ = θ / e12 :

cos γ
cos θ = = coshψ ≥ 1 e >1
sin γ
2
+ cos 2 γ
1’ e 2




Taking into account the relations between trigonometric and hyperbolic functions:

ψ ψ sinhψ
= coshψ =
cos sin
e12 e12 e12

the relation of the hyperbolic angle ψ with the Cartesian coordinates is:

x y
coshψ = sinhψ = e12
OQ OR


)
(
OP = ± (OQ coshψ + OR e12 sinhψ ) = ± OQ coshψ + e 2 ’ 1 sinhψ OQ e12

Let us define OS, which is taken
usually as the minor half-axis of the Figure 11.17
hyperbola although this definition is
not exact, as (figure 11.17):

OS = e 2 ’ 1 OQ e12

Then the equation of the hyperbola is:

OP = OQ coshψ + OS sinhψ

When the hyperbolic angle tends to
infinity we find the asymptotes:

OQ + OS
expψ
OP ’ ±
ψ’∞
2

OQ ’ OS
exp(’ ψ )
OP ’ ±
ψ’’∞
2

Two radii are conjugate if they are turned through the same hyperbolic angle • (figure
11.17):

OQ' = OQ cosh • + OS sinh•
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 131


OS' = OQ sinh• + OS cosh •

Then the equality OQ' 2 ’ OS' 2 = OQ2 ’ OS2 holds and the central radius of any point of
the hyperbola is:

OP = ± (OQ' cosh(ψ ’ • ) + OS' sinh(ψ ’ • ))

Also it is verified that the diameter is formed by the midpoints of the chords parallels to
the conjugate diameter.
A complete understanding of the central equation of the hyperbola is found in the
hyperbolic prism in a pseudo-Euclidean space. The equality OQ' 2 ’ OS' 2 = OQ2 ’ OS2 is
the condition of hyperbolic prism of constant radius. Then the hyperbolas are planar
section of this hyperbolic prism.


Conic passing through five points

Five non collinear points determine a conic. Since every conic has a Cartesian
second degree equation:

a x2 + b y2 + c x y + d x + e y + f = 0

where there are five independent parameters, the substitution of the coordinates of five
points leads to a linear system with five equation, with a hard solving. A briefer way to
obtain the Cartesian equation of the conic passing through these points is through the
Chasles™ theorem. Let us see an example:

C = (1, ’1) E = (’1, 3)
A = (1, 1) B = (2, 3) D = (0, 0)

EA = (2, ’2) EC = (2, ’4) ED = (1, ’3)
EB = (3, 0)

The projective cross ratio of the four points A, B, C and D on the conic is:

EA § EC EB § ED ’ 4 (’ 9 ) 3
r = {E , A B C D} = = =
EA § ED EB § EC ’ 4 (’ 12 ) 4

If the point X = (x, y) then:

XA = (1’x, 1’y) XB = (2’x, 3’y) XC = (1’x, ’1’y) XD = (’x, ’y)

According to the Chasles™ theorem, the projective cross ratio is constant for any point X
on the conic:

3 XA § XC XB § XD (2 x ’ 2 ) (3x ’ 2 y )
’ 0 = 12 x 2 ’ 3 y 2 ’ x y ’ 9 x + y
= =
4 XA § XD XB § XC ( x ’ y ) (4 x ’ y ’ 5)

Another way closest to geometric algebra is as follows. Let us take A, B and C as a
RAMON GONZALEZ CALVET
132

point base of the plane and express D and X in this base

D = d A A + d B B + dCC d A + d B + dC = 1
with

X = x A A + x B B + xC C x A + x B + xC = 1
with

The projective cross ratio is the quotient of the areas of the triangles XAC, XBD,
XAD and XBC:
x A x B xC x A x B xC
1 0 0 0 1 0
x B (x A d C ’ x C d A )
XA § XC XB § XD 0 0 1 dA dB dC
r= = =
(x B d C ’ xC d B ) x A
XA § XD XB § XC xA xB xC xA xB xC
1 0 0 0 1 0
dA dB dC 0 0 1

which yields the following equation:

0 = r x C x A d B + (1 ’ r ) x A x B d C ’ x B x C d A

The fifth point E also lying on the conic fulfils this equation:

0 = r eC e A d B + (1 ’ r ) e A e B d C ’ e B eC d A

which results in a simplified expression for the cross ratio:

d A dC

e eC
r= A
d B dC

e B eC

The substitution of the cross ratio r in the equation of the conic gives:

d A e A x B x C (d B eC ’ d C e B ) + d B e B x C x A (d C e A ’ d A eC ) + d C eC x A x B (d A e B ’ d B e A ) =

d A eA d B eB d C eC
d C xC = 0
d A xA d B xB
eA x A eB x B eC x C


Conic equations in barycentric coordinates and tangential conic

The Cartesian equation of a conic:

a x2 + b y2 + c x y + d x + e y + f = 0
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 133



is written in barycentric coordinates as a bilinear mapping:

« 
d e
+f +f
¬f ·
· «1 ’ x ’
2 2 y
¬
¬ ·
c+d +e
d
(1 ’ x ’ y y) ¬ + f + f ·¬
a+d + f ·=0
x x
¬2 ·¬
2 ·
y
¬e ·
c+d +e 
¬ +f +f b+e+ f ·
2 2 

Let us calculate now the dual conic of a given conic. This is defined as the locus of
the points which are dual of the lines tangent to the conic. Then the conic is the envelope
of the tangents. Let us differentiate the Cartesian equation:

δx (2 a x + c y + d ) + δy (2 b y + c x + e ) = 0

where δ indicates the ordinary differential in order to avoid confusion with the coefficient
d. The equation of the line touching the conic at (x0, y0) is:

(2 a x 0 + c y 0 + d ) (x ’ x 0 ) + (2 b y 0 + c x 0 + e )( y ’ y 0 ) = 0

(2 a x 0 + c y 0 + d ) x + (2 b y 0 + c x 0 + e ) y ’ 2 a x 02 ’ 2 b y 02 ’ 2 c x 0 y0 ’ d x0 ’ e y0 = 0

(2 a x 0 + c y 0 + d ) x + (2 b y 0 + c x 0 + e ) y + d x 0 + e y 0 + 2 f =0

(d x 0 + e y 0 + 2 f )(1 ’ x ’ y ) + ((2 a + d ) x 0 + (c + e ) y 0 + d + 2 f ) x +
+ ((2 b + e ) y 0 + (c + d ) x 0 + e + 2 f ) y = 0

If we denote by t, u and v the dual coordinates, the dual homogeneous coordinates t', u'
and v' are linear functions of the barycentric point coordinates:

d +2 f e+2 f  «1 ’ x 0 ’ y 0 
® t'  « 2 f
u'  = ¬ d + 2 f ·¬ ·
2a+2d +2 f c+d +e+2 f ·¬ x0
¬ ·
 v'  ¬ e + 2 f 2b+2e+2 f ·¬ ·
c+d +e+2 f y0
°»  

Now we see that this is twice the matrix of the conic equation (the matrix of a conic is
defined except by a common factor). Let M be the matrix of the conic, X the matrix of the
point coordinates, and U the matrix of the dual (homogeneous or normalised) coordinates.

U=MX

M ’1 U = X

XT = UT M ’1

because like M, the inverse matrix M ’1 is also symmetric and does not change under
RAMON GONZALEZ CALVET
134

transposition. The substitution in the equation of the conic XT M X = 0 gives:

UT M ’1 U = 0

That is, the matrix of the dual conic equation is the inverse of the matrix of the point
conic. Every tangent line of the conic is mapped onto a point of the dual conic and vice
versa. Let us consider for instance the ellipse:

«16 16 16  «1 ’ x ’ y
¬ ·¬ ·
2 2
«x « y
(1 ’ x ’ y y ) ¬16 15 16 · ¬
¬ · + ¬ · =1 ” ·=0
x x
4 2 ¬16 16 12 · ¬ ·
y
  

The inverse matrix is:
’1
« ’ 19 / 16 1
«16 16 16  1/ 4 
¬ · ¬ ·
=¬ 1 ’1
16 15 16 · 0·
¬
¬16 16 12 · ¬ 1/ 4 0 ’ 1/ 4·
   

So the equation of the dual conic is:

« ’ 19 16 4  ®t 
¬ ·
[ t u v ] ¬ 16 ’ 16 0 · u = 0

¬4 · v 
’ 4 ° »
0


where we can substitute indistinctly homogeneous or normalised coordinates. If we take t
= 1 ’ u ’ v then we obtain the Cartesian equation:

’ 67 u 2 ’ 31 v 2 ’ 78 u v + 70 u + 46 v ’ 19 = 0

A concrete case is the tangent line at (0, 2) with equation y = 2. The dual coordinates of
this line are obtained as follows:

® 2 1
[t' v' ] = [’ 2, ’ 2, ’ 1] =  , 
y ’ 2 ≡ t' (1 ’ x ’ y ) + u' x + v' y ’ u'
° 5 5»

Let us check that this dual point lies on the dual conic:

« ’ 19 16 4  ®2 
¬ ·
[2 2 1] ¬ 16 ’ 16 0 · 2 = 0

¬4 · 1 
’ 4 ° »
0



Polarities

A correlation is defined as the geometric transformation which maps collinear
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 135

points onto a pencil of lines, in other words, a linear mapping of points to lines (dual
points), which may be represented with a matrix M:

«1 ’ x ’
®a'  y
b'  = M ¬ ·
x
¬ ·

¬ ·
 c'  y
°»  

where a', b' and c' are homogeneous dual coordinates.
A polarity is a correlation whose square is the identity. This means that by
applying it twice, a point is mapped onto itself and a line is mapped onto itself. Let us
consider the line [a, b, c]:

«1 ’ x ’ y ®a' 
¬ · b'  = 0
[a b c] M ’1
[a b c] ¬ x ·=0 ’ 
¬ ·  c' 
y
  °»

On the other hand:

«1 ’ x' ’ y'  ®a 
¬ · b  = 0
[ a' c' ] ¬ [ a' c' ] M ’1
·=0
b' x' b'
’ 
¬ · c
y'
  °»

since [ a'' b'' c'' ]= k [ a b c ] where k is a homogeneous constant. Whence it follows
that the matrix M is symmetric for a polarity.
Under a polarity a point is mapped to its polar line. This point is also called the
pole of the line. The polarity transforms the pole into the polar and the polar into the pole.
In general the polar does not include the pole, except by a certain subset of points on the
plane. The set of points belonging to their own polar line (self-conjugate points) fulfil the
equation:
«1 ’ x ’ y 
¬ ·
(1 ’ x ’ y x y ) M ¬ x · = 0
¬ ·
y
 

This is just the equation of a point conic. The set of lines passing through their own poles
(self-conjugate lines) fulfil the equation:
Figure 11.18
®a 
[a b c] M ’1 b  = 0

c 
°»

which is just the equation of its dual
(tangential) conic. Then, a polarity has an
associated point and tangential conics, and
every conic defines a polarity. However
RAMON GONZALEZ CALVET
136

depending on the eigenvalues of the matrix, there are polarities that do not have any
associated conic.
The polar of a point outside its associated conic is the line passing through the
points of contact of the tangents drawn from the point (figure 11.18). The reason is the
following: the polarity maps the tangent lines of the associated conic to the tangency
points, so its intersection (the pole) must be transformed into the line passing through the
tangency points.


Reduction of the conic matrix to a diagonal form

Since the matrix of a polarity is symmetric, we can reduce it to a diagonal form.
Let D be the diagonal matrix and B the exchange base matrix. Then:

M = B ’1 D B

« d1 0
0
¬ ·
D=¬ 0 d2 0·
¬0 d3 ·
0
 

There exist three eigenvectors of the conic matrix. These eigenvectors are also
characteristic vectors of the dual conic matrix with inverse eigenvalues:

M ’1 = B ’1 D ’1 B

«1 
¬ 0·
0
¬ d1 ·
¬ ·
1 1
D ’1 M ’1 Vi =
=¬ 0 d i Vi = M Vi
0· Vi

d2 di
¬ ·

¬0 0
¬ d3 ·
 

The eigenvectors are three points (equal for both punctual and dual planes) that I will call
the eigenpoints of the polarity. Let X be a point on the conic given in Cartesian
coordinates. Then:

XT M X = XT B ’1 D B X = VT D V

where V = B X are the coordinates of the point in the base of eigenpoints. If we name with
t, u, v the eigencoordinates, then the equation of the conic is simply:

d1 t 2 + d 2 u 2 + d 3 v 2 = 0

The eigenpoints of a polarity do not never belong to the associated conic since they have
the eigencoordinates (1,0,0), (0,1,0) and (0,0,1) and di ≠ 0. On the other hand, the conic
only exist if the eigenvalues have different signs (one positive and two negative or one
negative and two positive). This fact classifies the polarities in two classes: those having
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 137

an associated conic, and those not having an associated conic. In fact, if we state a level, a
certain value of the matrix product, there can be a conic or not. That is, there is a family of
conics for different levels and there is a threshold value, from where the conic does not
exist.


Using a base of points on the conic

If instead of Cartesian coordinates we use the coordinates of a base of three points
A, B and C lying on the conic, the coefficients on the matrix diagonal vanish. Then the
equation of the conic is:

«0 c13  « x A 
c12
¬ ·¬ ·
1
(x A x C ) ¬ c12 c 23 · ¬ x B · = c 23 x B x C + c 31 x C x A + c12 x A x B = 0
xB 0
2 ¬c 0 · ¬ xC ·
c 23
 13  

With the same base, the equation of the conic passing through D and whose projective
cross ratio {X, A B C D} is r is:

0 = r x C x A d B + (1 ’ r ) x A x B d C ’ x B x C d A

From where we have:

(1 ’ r ) d C
« r dB  « xA 
0
¬ ·¬ ·
(x A x C ) ¬ (1 ’ r ) d C ’ d A · ¬ xB · = 0
xB 0
¬ rd 0 · ¬ xC ·
’ dA
  
B



This conic matrix is not constant, and has the determinant:

det = ’2 r (1 ’ r ) d A d B d C

so the matrix with constant coefficients is:

(1 ’ r ) d C
« r dB 
0
¬ ·
1
C= ¬ (1 ’ r ) d C ’ dA ·
0
3 ’ r ( ’ r) d d d
1 C¬

’ dA
 r dB
A B



Exercises

11.1 Prove that the projective cross ratio of any four points on a conic is equal to the ratio
of the sines of the cylindrical half-angles. That is, if:

OA = OQ cos ± + OR sin ± OB = OQ cos β + OR sin β , etc.
RAMON GONZALEZ CALVET
138

γ ’± δ ’β
sin sin
2 2
Then { X , A B C D} =
δ ’± γ ’β
sin sin
2 2

11.2 Prove that an affinity transforms a circle into an ellipse, and using this fact prove
the Newton™s theorem: Given two directions, the ratio of the product of distances from
any point on the plane to both intersections with an ellipse along the first direction and
the same product of distances along the other direction is independent of the location of
this point on the plane.

11.3 Prove that a line touching an ellipse is parallel to the conjugate diameter of the
diameter whose end-point is the tangency point. Prove also that the parallelogram
circumscribed around an ellipse has constant area (Apollonius™ theorem).

11.4 Prove the former theorem applied to a hyperbola: the parallelogram formed by two
conjugate diameters has constant area.

11.5 Given a pole, trace a line passing through the pole and cutting the conic. Prove that
the pole, and the intersections of this line with the conic and with the polar form a
harmonic range.

11.6 Calculate the equation of the point and tangential conic passing through the points
(1, 1), (2, 3), (1, 4), (0, 2) and (2, 5).

11.7 The Pascal™s theorem: Let A, B, C, D, E and F be six distinct points on a proper
conic, Let P be the intersection of the line AE with the line BF, Q the intersection of the
line AD with CF, and R the intersection of the line BD with CE. Show that P, Q and R are
collinear.

11.8 The Brianchon™s theorem. Prove that the diagonals joining opposite vertices of a
hexagon circumscribed around a proper conic are concurrent. Hint: take the dual plane.
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 139


THIRD PART: PSEUDO-EUCLIDEAN GEOMETRY

This part is devoted to the hyperbolic plane, where vectors and numbers have a
pseudo Euclidean modulus, that is, a modulus of a Minkowski space. The bidimensional
geometric algebra already includes the Euclidean and pseudo Euclidean planes. In fact,
the geometric algebra does not make any special distinction between both kinds of
planes. On the other hand, the plane geometric algebra can be represented by real 2—2
matrices, which helps us to define some concepts with more precision.

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