ñòð. 6 |

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

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

Extracting common factor:

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

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

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

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

=

E'A' âˆ§ E'D' E'B' âˆ§ E'C' EA âˆ§ ED EB âˆ§ 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 Chaslesâ€™ 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 Ptolemyâ€™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 âˆ’ 2 AC âˆ’1 BC ) âˆ’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' âˆ§ HC' HB' âˆ§ HD' HA âˆ§ HC HB âˆ§ HD

=

HA' âˆ§ HD' HB' âˆ§ HC' HA âˆ§ HD HB âˆ§ 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' âˆ’1 = ( OP âˆ’1 + v ) âˆ’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:

â†’

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: ï£¦PFï£¦=ï£¦PSï£¦. Also PF' and PS' are tangent to

the other sphere so that their lengths are also equal: ï£¦PF'ï£¦=ï£¦PS'ï£¦. 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 ï£¦PFï£¦=ï£¦PSï£¦. But ï£¦PSï£¦

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):

sinÎ±

d (P , r )

PF = PS =

sinÎ´

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

sinÎ±

e= >0 Figure 11.2

sinÎ´

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 )

ï£¦FPï£¦ 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 Â· 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 Â· PT

FP = e d (P, r ) = e

FT

FP FT = e FT Â· PT = e FT Â· (FT âˆ’ FP ) = e (FT 2 âˆ’ FT Â· FP )

Let us denote by Î² the angle between the main axis FT and the focal vector FP:

FP FT = e (FT 2 âˆ’ FT FP cos Î² )

FP FT (1 + e cos Î² ) = e FT 2

From here one obtains the polar equation of a conic:

FT

FP =

1 + e cos Î²

According to the value of the eccentricity there are three types of proper conics:

1) For 0 < e < 1, the inclination Î± of the plane of the conic is lower than the

inclination Î´ 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, Î± = Î´ , the plane of the conic is parallel to a cone generatrix. Since

the denominator vanishes for Î² = Ï€, the points P form an open curve called

parabola. Except for this value of Î², the radius is always positive and the

curve lies on the focal half-plane.

3) For e > 1, Î± >Î´ , 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 Î²a :

ï£« 1ï£¶

Î² a = arccosï£¬ âˆ’ ï£·

ï£ eï£¸

The maximum eccentricity of the

hyperbolas as cone sections is obtained

with Î± =Ï€/2:

RAMON GONZALEZ CALVET

120

1

1< e â‰¤

sinÎ´

The focal radius is positive (P lies on the focal branch) for the ranges of Î² :

0 < Î² < Î²a 2Ï€ â€“ Î²a < Î² < 2Ï€

and

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

Î²a < Î² < 2Ï€ â€“ Î²a

A negative focal radius means that we must take the sense opposite that determined by the

angle (figure 11.4). For example, a radius â€“2 with an angle 7Ï€/6 is equivalent to a radius 2

with an angle of Ï€/6. So the vertex T' of the non focal branch has an angle Ï€ (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 ï£¦FQï£¦ instead of the distance from the focus to the directrix ï£¦FTï£¦:

1+ e

FP = FQ

1 + e cos Î²

Let Q' be the vertex of the conic closest to the focus F'. The distance from the focus F to

Q' is found for Î² = Ï€:

1+ e

FQ' = FQ

1âˆ’ e

For the ellipse ï£¦FQ'ï£¦ 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' âˆ’ F'Q' = FQ' âˆ’ FQ = FQ

1âˆ’ e

On the other hand the major diameter is:

2 FQ

QQ' = FQ + FQ' =

1âˆ’ 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 Î² (figure 11.3), FP is obtained

from the unitary vector of FQ via multiplication by the exponential with argument Î² and

by the modulus of FP yielding:

1+ e

FQ (cos Î² + e12 sinÎ² )

FP =

1 + e cos Î²

On the other hand, from the directrix property, one easily finds the following

equation for a conic:

FP2 FT2 = e2 ( FT2 âˆ’ FT Â· 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 Â½ 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 âˆ’ 3 ) e1 + ( y âˆ’ 4 ) e2

Then the equation of the conic is:

1

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

4

After simplification:

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

The Chaslesâ€™ 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, TraitÃ© 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 = ï£¦FQï£¦ e1

Now on Î±, Î², Î³, Î´ and Ï‡ 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:

ï£® e cos Î± + e 2 sinÎ± e1 cos Ï‡ + e 2 sinÏ‡ ï£¹

XA = FA âˆ’ FX = FQ (1 + e ) ï£¯ 1 âˆ’ ï£º

1 + e cos Î± 1 + e cos Ï‡

ï£° ï£»

Introducing a common denominator, we find:

(1 + e ) [e1 (cos Î± âˆ’ cos Ï‡ ) + e 2 (sinÎ± âˆ’ sinÏ‡ + e sinÎ± cos Ï‡ âˆ’ e cos Î± sinÏ‡ )]

XA = FQ

(1 + e cos Î± ) (1 + e cos Ï‡ )

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

e (1 + e ) (sinÎ³ cos Î± âˆ’ sinÎ± cos Î³ + sinÏ‡ cos Î³ âˆ’ sinÎ³ cos Ï‡ + sinÎ± cos Ï‡ âˆ’ sinÏ‡ cos Î± )

2

XA âˆ§ XC = FQ 122

(1 + e cos Î± )(1 + e cos Î³ )(1 + e cos Ï‡ )

(1 + e )2 [ sin(Î³ âˆ’ Î± ) + sin (Ï‡ âˆ’ Î³ ) + sin (Î± - Ï‡ ) ]

= FQ 2

e12

(1 + e cos Î± )(1 + e cos Î³ )(1 + e cos Ï‡ )

Using the trigonometric identities of the half-angles, the sum is converted into a product

of sines (exercise 6.2):

( 1 + e )2 ï£® sinï£« Î³ âˆ’ Î± ï£¶ sinï£« Ï‡ âˆ’ Î³ ï£¶ sinï£« Î± âˆ’ Ï‡ ï£¶ ï£¹

ï£¯ ï£¬ 2 ï£· ï£¬ 2 ï£· ï£¬ 2 ï£·ï£º

ï£°ï£ ï£¸ï£ ï£¸ï£ ï£¸ï£»

XA âˆ§ XC = âˆ’ 4 FQ 2

e12

( 1 + e cos Î± ) ( 1 + e cos Î³ ) ( 1 + e cos Ï‡ )

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 Ï‡ are simplified:

RAMON GONZALEZ CALVET

124

Î³ âˆ’Î± Î´ âˆ’Î² AFC BFD

sin sin sin sin

XA âˆ§ XC XB âˆ§ XD 2 2= 2 2

{X , A B C D } = =

Î´ âˆ’Î± Î³ âˆ’Î²

XA âˆ§ XD XB âˆ§ XC AFD BFC

sin sin sin sin

2 2 2 2

since Î³ âˆ’Î± 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

Chaslesâ€™ 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 Î± + e 2 sinÎ± )

FP =

1 + e cosÎ±

The derivation with respect the angle Î± gives:

( 1 + e ) FQ

d FP

[âˆ’ e sinÎ± + e2 ( e + cos Î± ) ]

=

( 1 + e cos Î± )2 1

dÎ±

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

vector t being:

âˆ’ e1 sinÎ± + e 2 ( e + cos Î± )

t=

1 + e 2 + 2 e cos Î±

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

e1 ( e + cos Î± ) + e 2 sinÎ±

n=

1 + e 2 + 2 e cos Î±

In the same way, t and n are obtained as functions of the angle Î² (figure 11.9) from the

vectorial equation for the focus F':

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 125

(1 âˆ’ e ) FQ

( e1 cos Î² + e 2 sinÎ² )

F'P =

1 âˆ’ e cosÎ²

âˆ’ e1 sinÎ² + e 2 (âˆ’ e + cos Î² )

t=

1 + e 2 âˆ’ 2 e cos Î² Figure 11.10

e1 ( âˆ’ e + cos Î² ) + e 2 sinÎ²

n=

1 + e 2 âˆ’ 2 e cos Î²

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:

ï£¦FPï£¦ + ï£¦F'Pï£¦ = constant â‡’ 0 = dï£¦FPï£¦ + dï£¦F'Pï£¦

The differential of the focal

4

Figure 11.11

radius is obtained by differentiating the

square of the focal vector:

d FP2 = 2 FP Â· d FP = 2 ï£¦FPï£¦ dï£¦FPï£¦

d FP Â· FP

d FP = âˆ’

FP

Summing the differentials of both

moduli we obtain:

ï£« FP F'P ï£¶

d FP Â· FP d F'P Â· F'P

= âˆ’ d FP Â· ï£¬ ï£·

0 = d FP + d F'P = âˆ’ âˆ’ +

ï£¬ FP F'P ï£·

FP F'P ï£ ï£¸

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 dï£¦FPï£¦ and dï£¦F'Pï£¦ 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: dï£¦FPï£¦â‰ ï£¦d FPï£¦.

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= 1âˆ’ e = Figure 11.14

OQ OQ

For the case of the ellipse all the quantities are positive. For the case of the

hyperbola both ï£¦OFï£¦ and ï£¦OQï£¦ are negative with ï£¦FQï£¦ positive. Let the angle QOP be

Î³ (figure 11.14). Then:

ï£¦OPï£¦ sin Î³ = ï£¦FPï£¦ sin Î±

ï£¦OPï£¦ cos Î³ = ï£¦OFï£¦ + ï£¦FPï£¦ cos Î±

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 127

For the ellipse all the moduli are positive whereas for the hyperbola ï£¦OPï£¦ and

ï£¦OFï£¦ 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 âˆ’ e )sinÎ±

(1 + e ) sinÎ± 2

OP sinÎ³ = FQ = OQ

1 + e cos Î± 1 + e cos Î±

(1 âˆ’ e 2 )cos Î± ï£¹ = OQ e + cosÎ±

FQ (1 + e ) cosÎ± ï£®

OP cosÎ³ = OF + = OQ ï£¯e + ï£º

1 + e cos Î± 1 + e cos Î± ï£» 1 + e cos Î±

ï£°

Summing the squares of both former equalities we find:

ï£« sin 2 Î³ ï£¶ OQ

+ cos 2 Î³ ï£· = OQ 2

OP ï£¬ OP =

2

â‡’

ï£¬1 âˆ’ e2 ï£·

sin 2 Î³

ï£ ï£¸

+ cos 2 Î³

1âˆ’ e 2

The direction of OP with respect to OQ is given by the angle Î³ and we may add it

multiplying by the unitary complex with argument Î³ to obtain the central vectorial

equation:

OQ (cos Î³ + e12 sinÎ³ )

OP =

sin 2 Î³

+ cos 2 Î³

1âˆ’ e 2

Let R be the point P for Î³ =Ï€/2. Then OR lies on the secondary axis of symmetry

of the conic, which is perpendicular to OQ, the main axis of symmetry. ï£¦ORï£¦is the minor

half-axis. The eccentricity relates both:

OR2 = ( 1 âˆ’ e2 ) OQ2

OR = 1 âˆ’ e 2 OQ

OR = 1 âˆ’ e 2 OQ e12

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

ï£« sinÎ³ ï£¶

1

ï£¬ OQ cos Î³ + OR ï£·

OP = ï£¬ 2ï£·

sin 2 Î³ 1âˆ’ e ï£¸

ï£

+ cos 2 Î³

1âˆ’ e 2

Hence it follows:

OP 2 OP 2

sin Î³ + cos 2 Î³ = 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 = ï£¦OPï£¦ cos Î³ y = ï£¦OPï£¦ sin Î³

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 ï£¦OQï£¦ is imaginary and

OQ2<0 converting the sum of the squares in the former equation in a difference of real

positive quantities.

Diameters and Apolloniusâ€™ theorem

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

Figure 11.15

cos Î³

cos Î¸ =

sin 2 Î³

+ cos 2 Î³

1âˆ’ e 2

Then the central equation of an ellipse

becomes5:

OP = OQ cos Î¸ + OR sin Î¸

The angle Î¸ has a direct geometric

interpretation if we look at a section of a

cylinder (figure 11.15):

cos Î³ cos Ï†

cos Î¸ =

sin 2 Î³ + cos 2 Î³ cos 2 Ï†

From where it follows the relationship between the inclination of the cylindrical section

and the eccentricity:

cos Ï† = 1 âˆ’ 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 Î¸. 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 Î¸ 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 Ï‡ (figure 11.16):

OQ' = OQ cos Ï‡ + OR sinÏ‡ OR' = âˆ’OQ sin Ï‡ + OR sinÏ‡

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 Ï‡ âˆ’ OR' sin Ï‡

Figure 11.16

OR = OQ' sin Ï‡ + OR' cos Ï‡

Then, the central radius of any point P is:

OP = ( OQ' cos Ï‡ âˆ’ OR' sin Ï‡ ) cos Î¸ + ( OQ' sin Ï‡ + OR' cos Ï‡ ) sin Î¸

= OQ' cos( Î¸ âˆ’Ï‡ ) + OR' sin( Î¸ âˆ’Ï‡ )

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(Î¸ âˆ’Ï‡ ) âˆ’ OR' sin(Î¸ âˆ’Ï‡ ) = OQ' cos(Ï‡ âˆ’Î¸ ) + OR' sin(Ï‡ âˆ’Î¸ )

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

PP' = OP' âˆ’ OP = âˆ’2 OR' sin( Î¸ âˆ’ Ï‡ )

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(Î¸ âˆ’ Ï‡ )

2

The properties of the conjugate diameters are also applicable to the hyperbola, but

taking into account that OR2 = (1 âˆ’ e2) OQ2 < 0, that is, the minor half-axis ï£¦ORï£¦ is an

imaginary number. In this case OR has the same direction than OQ:

RAMON GONZALEZ CALVET

130

OR = 1 âˆ’ e 2 OQ e12 = e 2 âˆ’ 1 OQ

And also the angle Î¸ is an imaginary number, and the trigonometric functions are turned

into hyperbolic functions of the real argument Ïˆ = Î¸ / e12 :

cos Î³

cos Î¸ = = coshÏˆ â‰¥ 1 e >1

sin Î³

2

+ cos 2 Î³

1âˆ’ e 2

Taking into account the relations between trigonometric and hyperbolic functions:

Ïˆ Ïˆ sinhÏˆ

= coshÏˆ =

cos sin

e12 e12 e12

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

x y

coshÏˆ = sinhÏˆ = e12

OQ OR

)

(

OP = Â± (OQ coshÏˆ + OR e12 sinhÏˆ ) = Â± OQ coshÏˆ + e 2 âˆ’ 1 sinhÏˆ 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 âˆ’ 1 OQ e12

Then the equation of the hyperbola is:

OP = OQ coshÏˆ + OS sinhÏˆ

When the hyperbolic angle tends to

infinity we find the asymptotes:

OQ + OS

expÏˆ

OP â†’ Â±

Ïˆâ†’âˆž

2

OQ âˆ’ OS

exp(âˆ’ Ïˆ )

OP â†’ Â±

Ïˆâ†’âˆ’âˆž

2

Two radii are conjugate if they are turned through the same hyperbolic angle Ï• (figure

11.17):

OQ' = OQ cosh Ï• + OS sinhÏ•

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 131

OS' = OQ sinhÏ• + OS cosh Ï•

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

the hyperbola is:

OP = Â± (OQ' cosh(Ïˆ âˆ’ Ï• ) + OS' sinh(Ïˆ âˆ’ Ï• ))

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 âˆ’ OS' 2 = OQ2 âˆ’ 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

Chaslesâ€™ theorem. Let us see an example:

C = (1, âˆ’1) E = (âˆ’1, 3)

A = (1, 1) B = (2, 3) D = (0, 0)

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

EB = (3, 0)

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

EA âˆ§ EC EB âˆ§ ED âˆ’ 4 (âˆ’ 9 ) 3

r = {E , A B C D} = = =

EA âˆ§ ED EB âˆ§ EC âˆ’ 4 (âˆ’ 12 ) 4

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

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

According to the Chaslesâ€™ theorem, the projective cross ratio is constant for any point X

on the conic:

3 XA âˆ§ XC XB âˆ§ XD (2 x âˆ’ 2 ) (3x âˆ’ 2 y )

â‡’ 0 = 12 x 2 âˆ’ 3 y 2 âˆ’ x y âˆ’ 9 x + y

= =

4 XA âˆ§ XD XB âˆ§ XC ( x âˆ’ y ) (4 x âˆ’ y âˆ’ 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 âˆ’ x C d A )

XA âˆ§ XC XB âˆ§ XD 0 0 1 dA dB dC

r= = =

(x B d C âˆ’ xC d B ) x A

XA âˆ§ XD XB âˆ§ 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 âˆ’ r ) x A x B d C âˆ’ 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 âˆ’ r ) e A e B d C âˆ’ e B eC d A

which results in a simplified expression for the cross ratio:

d A dC

âˆ’

e eC

r= A

d B dC

âˆ’

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 âˆ’ d C e B ) + d B e B x C x A (d C e A âˆ’ d A eC ) + d C eC x A x B (d A e B âˆ’ 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:

ï£« ï£¶

d e

+f +f

ï£¬f ï£·

ï£· ï£«1 âˆ’ x âˆ’

2 2 yï£¶

ï£¬

ï£¬ ï£·

c+d +e

d

(1 âˆ’ x âˆ’ y y) ï£¬ + f + f ï£·ï£¬

a+d + f ï£·=0

x x

ï£¬2 ï£·ï£¬

2 ï£·

y

ï£¬e ï£·ï£

c+d +e ï£¸

ï£¬ +f +f b+e+ f ï£·

ï£2 2 ï£¸

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:

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

where Î´ 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 âˆ’ x 0 ) + (2 b y 0 + c x 0 + e )( y âˆ’ y 0 ) = 0

(2 a x 0 + c y 0 + d ) x + (2 b y 0 + c x 0 + e ) y âˆ’ 2 a x 02 âˆ’ 2 b y 02 âˆ’ 2 c x 0 y0 âˆ’ d x0 âˆ’ 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 âˆ’ x âˆ’ 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 ï£¶ ï£«1 âˆ’ x 0 âˆ’ y 0 ï£¶

ï£® t' ï£¹ ï£« 2 f

ï£¯u' ï£º = ï£¬ d + 2 f ï£·ï£¬ ï£·

2a+2d +2 f c+d +e+2 f ï£·ï£¬ x0

ï£¯ï£ºï£¬ ï£·

ï£¯ v' ï£º ï£¬ e + 2 f 2b+2e+2 f ï£·ï£¬ ï£·

c+d +e+2 f y0

ï£°ï£»ï£ ï£¸ï£ ï£¸

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 âˆ’1 U = X

XT = UT M âˆ’1

because like M, the inverse matrix M âˆ’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 âˆ’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:

ï£«16 16 16 ï£¶ ï£«1 âˆ’ x âˆ’ yï£¶

ï£¬ ï£·ï£¬ ï£·

2 2

ï£«xï£¶ ï£« yï£¶

(1 âˆ’ x âˆ’ y y ) ï£¬16 15 16 ï£· ï£¬

ï£¬ ï£· + ï£¬ ï£· =1 â‡” ï£·=0

x x

ï£4ï£¸ ï£2ï£¸ ï£¬16 16 12 ï£· ï£¬ ï£·

y

ï£ ï£¸ï£ ï£¸

The inverse matrix is:

âˆ’1

ï£« âˆ’ 19 / 16 1

ï£«16 16 16 ï£¶ 1/ 4 ï£¶

ï£¬ ï£· ï£¬ ï£·

=ï£¬ 1 âˆ’1

16 15 16 ï£· 0ï£·

ï£¬

ï£¬16 16 12 ï£· ï£¬ 1/ 4 0 âˆ’ 1/ 4ï£·

ï£ ï£¸ ï£ ï£¸

So the equation of the dual conic is:

ï£« âˆ’ 19 16 4 ï£¶ ï£®t ï£¹

ï£¬ ï£·

[ t u v ] ï£¬ 16 âˆ’ 16 0 ï£· ï£¯uï£º = 0

ï£¯ï£º

ï£¬4 ï£· ï£¯v ï£º

âˆ’ 4ï£¸ ï£° ï£»

0

ï£

where we can substitute indistinctly homogeneous or normalised coordinates. If we take t

= 1 âˆ’ u âˆ’ v then we obtain the Cartesian equation:

âˆ’ 67 u 2 âˆ’ 31 v 2 âˆ’ 78 u v + 70 u + 46 v âˆ’ 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:

ï£® 2 1ï£¹

[t' v' ] = [âˆ’ 2, âˆ’ 2, âˆ’ 1] = ï£¯ , ï£º

y âˆ’ 2 â‰¡ t' (1 âˆ’ x âˆ’ y ) + u' x + v' y â‡’ u'

ï£° 5 5ï£»

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

ï£« âˆ’ 19 16 4 ï£¶ ï£®2 ï£¹

ï£¬ ï£·

[2 2 1] ï£¬ 16 âˆ’ 16 0 ï£· ï£¯2ï£º = 0

ï£¯ï£º

ï£¬4 ï£· ï£¯1 ï£º

âˆ’ 4ï£¸ ï£° ï£»

0

ï£

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:

ï£«1 âˆ’ x âˆ’

ï£®a' ï£¹ yï£¶

ï£¯b' ï£º = M ï£¬ ï£·

x

ï£¬ ï£·

ï£¯ï£º

ï£¬ ï£·

ï£¯ c' ï£º y

ï£°ï£» ï£ ï£¸

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]:

ï£«1 âˆ’ x âˆ’ yï£¶ ï£®a' ï£¹

ï£¬ ï£· ï£¯b' ï£º = 0

[a b c] M âˆ’1

[a b c] ï£¬ x ï£·=0 â‡’ ï£¯ï£º

ï£¬ ï£· ï£¯ c' ï£º

y

ï£ ï£¸ ï£°ï£»

On the other hand:

ï£«1 âˆ’ x' âˆ’ y' ï£¶ ï£®a ï£¹

ï£¬ ï£· ï£¯b ï£º = 0

[ a' c' ] ï£¬ [ a' c' ] M âˆ’1

ï£·=0

b' x' b'

â‡’ ï£¯ï£º

ï£¬ ï£· ï£¯cï£º

y'

ï£ ï£¸ ï£°ï£»

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:

ï£«1 âˆ’ x âˆ’ y ï£¶

ï£¬ ï£·

(1 âˆ’ x âˆ’ y x y ) M ï£¬ x ï£· = 0

ï£¬ ï£·

y

ï£ ï£¸

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

ï£®a ï£¹

[a b c] M âˆ’1 ï£¯b ï£º = 0

ï£¯ï£º

ï£¯c ï£º

ï£°ï£»

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 âˆ’1 D B

ï£« d1 0ï£¶

0

ï£¬ ï£·

D=ï£¬ 0 d2 0ï£·

ï£¬0 d3 ï£·

0

ï£ ï£¸

There exist three eigenvectors of the conic matrix. These eigenvectors are also

characteristic vectors of the dual conic matrix with inverse eigenvalues:

M âˆ’1 = B âˆ’1 D âˆ’1 B

ï£«1 ï£¶

ï£¬ 0ï£·

0

ï£¬ d1 ï£·

ï£¬ ï£·

1 1

D âˆ’1 M âˆ’1 Vi =

=ï£¬ 0 d i Vi = M Vi

0ï£· Vi

â‡”

d2 di

ï£¬ ï£·

1ï£·

ï£¬0 0

ï£¬ d3 ï£·

ï£ ï£¸

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 âˆ’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 â‰ 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:

ï£«0 c13 ï£¶ ï£« x A ï£¶

c12

ï£¬ ï£·ï£¬ ï£·

1

(x A x C ) ï£¬ c12 c 23 ï£· ï£¬ x B ï£· = c 23 x B x C + c 31 x C x A + c12 x A x B = 0

xB 0

2 ï£¬c 0 ï£· ï£¬ xC ï£·

c 23

ï£ 13 ï£¸ï£ ï£¸

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 âˆ’ r ) x A x B d C âˆ’ x B x C d A

From where we have:

(1 âˆ’ r ) d C

ï£« r dB ï£¶ ï£« xA ï£¶

0

ï£¬ ï£·ï£¬ ï£·

(x A x C ) ï£¬ (1 âˆ’ r ) d C âˆ’ d A ï£· ï£¬ xB ï£· = 0

xB 0

ï£¬ rd 0 ï£· ï£¬ xC ï£·

âˆ’ dA

ï£ ï£¸ï£ ï£¸

B

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

det = âˆ’2 r (1 âˆ’ r ) d A d B d C

so the matrix with constant coefficients is:

(1 âˆ’ r ) d C

ï£« r dB ï£¶

0

ï£¬ ï£·

1

C= ï£¬ (1 âˆ’ r ) d C âˆ’ dA ï£·

0

3 âˆ’ r ( âˆ’ r) d d d

1 Cï£¬

0ï£·

âˆ’ dA

ï£ r dB

A B

ï£¸

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 Î± + OR sin Î± OB = OQ cos Î² + OR sin Î² , etc.

RAMON GONZALEZ CALVET

138

Î³ âˆ’Î± Î´ âˆ’Î²

sin sin

2 2

Then { X , A B C D} =

Î´ âˆ’Î± Î³ âˆ’Î²

sin sin

2 2

11.2 Prove that an affinity transforms a circle into an ellipse, and using this fact prove

the Newtonâ€™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 (Apolloniusâ€™ 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 Pascalâ€™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 Brianchonâ€™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 2Ã—2

matrices, which helps us to define some concepts with more precision.

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