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Conics - a class of plain curves

Mathematic

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C O N I C S

Conics reprezent a class of plain curves with special properites, which have numerous applications in various domains.



The mathematicians of the elenic era obtained non-degenerated conics by the intersection of a rotation conic with a plane. The books of Apollonius (262-200 BC) describe for the first time the elypse, hyperbola, parabola and their properties.

We will make a short introduction into the world of conics by using the reduced equation, then we will classify izometracally the curves of 2nd order. Using the knowledge from the previous chapters, we will understand that the geometrical locus of points in the euclidian plane, in an othonormed frame is the intersection of a circular cone with a plane, or Null.

The general properties of conicsm and the conics with initial conditions, will be studied in the final part of this chapter.

1.Conics given by reduced equations

Let be E2 the bidimensional punctual euclidian space, and

R (O,) a carthesian orthonormed frame. A plane curve is determined in a carthesian refrence frame by an equation with 2 variables: F ( x,y ) = 0. A particular case of the plane curves are the conics. Behold, the geometrical definitions and the algebrical properties of conics in a freely chosen refrence frame:

1.1 Elypse: The geometrical locus of points belonging to the euclidian plane with the property that the sum of distances from 2 fixed points F1 and F2 stays constant.

Given xOy is an orthogonal reference frame in the euclidian plane E2 aI R and the points F1 (-c,0) , F2 (c,0) are fixed, then the all points M(x,y) I E2 with the property MF1 + MF2 = 2a are algebrically written as follows:

, c = (1.1)

y

B M(x,y)

A Ooo A x

x = - B x =

fig.1

For the elypse (1.1) following notions are to be considered: (fig.1) :

  • F1, F2 are the focal points of the elypse, and F1F2 = 2 c focal distance
  • a greater semiaxis, iar b- small semiaxis
  • A(a,0) , A(-a,0) , B(b,0), B(-b,0) edges of the elypse
  • The lines x = , are the directing lines of the elypse
  • e = < 1 - excentricity of the elypse

The Axes Ox ,Oy of the carthesian frame are axes of symmetry, the origin of the frame is the center of the elypse. Therefore the orthonormed frame R(O, ) is also called canonical and the equation (1.1) is called reduced.

The elypse (1.1) is the geometrical locus of the points M(x,y) with satisfy one of the following properties:

or

The elypse of with the semiaxis a and b respectively, can be parametrically written as: x = a cosj , y = b sin j , j I [0,2p] .

It can be easily proven that the perpendicular on the tangent of any point of the elypse is the bisecting line of the angle between the focal radiuses to that point. (The optical property of the elypse).

1.2 Hyperbola : The geometrical locus of the points from the euclidian plane E2 for which the modulus of the diffrence of distances of two distinct, fixed points is constant.

Given xOy an orthogonal refrence frame in the euclidian plane E2, aI R and the fixed points F1 (-c,0) , F2 (c,0) then the points M(x,y) I E2 with the property MF1 - MF2 = 2a is algebrically written as:

, c = (1.2)

y


F1 A O A F2

Fig.2

  • F1(-c,o), F2 (c,o) focal points of the hyperbola
  • A(-a,0) , A(a,0) - edges of the hyperbola
  • a ,b - semiaxes of the hyperbola
  • the lines y = x are the asimptotes of the hyperbola which geometrically represent the diagonals of the square with the sides 2a and 2b respectively, with the center in O, the sides being parallel to the axes of symmetry.
  • The lines x = are the directing lines of the hyperbola
  • e = > 1 - excentricity of the hyperbola

The axes Ox , Oy of the frame R(O, ) are the symmetry axes of the hyperbola, and the origin of the frame is a symmetry center of the hyperbola, hence equation (1.2) is the reduced form of the hyperbola.

The Hyperbola characterized by the equation (1.2) is the geometrical locus of the points M(x,y) I E2 , which satisfy one of the following conditions:

or ,

where the lines d1 and d2 are the directing lines of the hyperbola.

The parametric equations of the hyperbola are given by:

x = a ch t , y = b sh t , tI R

The tangent to a point of the hyperbola, is the bisecting line of the angle between the focal radiuses. ( the optical property of the hyperbola )

1.3 Parabola : the geometric locus of points equally distanced from a fixed point F (focal) and a fixed line D (directing line) .

If xOy is an orthogonal refrence frame in the euclidian plane E2 , pI R + , and given the point F (,0) and the line (D): x = - then the set of points M(x,y) withthe property (M,F) = (M, D) is algebrically written as follows:

y2 = 2px , ( p > 0 ) (1.3)


D y

M(x,y)

O F(,0) x

Fig.3

We define the following notions associated to a parabola:

   F(,0) focal point of the parabola, the quantity focal distance

   O(0,0) - edge of the parabola

   Ox - transversal axis of the parabola, the symmetry axis

   Oy - tangent axis of the parabola

   the line D of equation x = - , is the directing line

   the excentricity of the parabola is e=1.

2.Second order curves in the euclidian plane

In the euclidian punctual space, referred to an affine reference frame the polinomial function (affine form) f :R2 R , is given by:

f (x,y) = a11x2 + 2a12xy + a22y2 + 2a13x + 2a23y + a33 , a112+ a122+ a2220.

Finally we will proove, that the set of points of the plane E2 , whose coordinates (x,y) are the roots of the function f, represent a conic, or a Null space, from the geometrical point of view, argumenting the following definition:

Definition. 2.1

The algebrical curve of second order, which includes the set of points of the plane E2 is called a conic, if its coordinates satisfy the following equation:

a11x2 + 2a12xy + a22y2 + 2a13x + 2a23y + a33 = 0 , a112+ a122+ a2220. (2.1)

If we consider the homogenous coordinates (x1,x2,x3) of a point M(x,y) , tied to the coordinates (x,y), called non-homogenous coordinates, trough the relations: x = , y = , we obtain:

f(x1,x2,x3) s a11x 12 + 2a12x1x2 + a22x222 + 2a13x1x3 + 2a23x2x3 + a33x32 = 0

or , where aij = aji (2.2)

The symmetrical matrix A = (aij) will be called the matrix of the conic ().

Let us consider the numbers:

D = det.A , , I = a11 + a22 .

If we apply a (izometrical) orthogonal transformation T : V2 V2 , T () = , T()=, in the euclidian punctual space E2, passing from the orthonormed frame R (O,) to the new orthonormed frame

R (O,) the conic () will have the following analytic equation:

, where D , d, I are calculated in the new frame.

Theorem 2.2

The quatities D , d , I are the orthogonal invariants of the conic ( g ), hence : D = D, d= d , I= I .

These invatriants will be called as follows: D-cubic invariant,

d - square invariant, I linear invariant .

Proof. We will demonstrate first the invariance of d and I. Let us consider the quadratic form:

j (x,y) = a11x2 + 2a12xy + a22y2 (2.3)

The symmetrical bilinear form, associated to the quadratic form j, has a unique symmetrical linear transformation T : E2 E2 having its associated matrix identical to that of the quadratic form (2.3). Hence the characteristic equation of the transformation T

l2 I l +d = 0,

is invariant to a base change, therefore d= d , I= I .

Writing the equation of the conic () with the homogenous coordinates (2.2) and repeating the procedure described above for the quadratic form f(x1,x2,x3), we can proove the invariance of the characteristic equation:

l3- J1l2 + J2l - D = 0,

therefore D is also invariant.

Definition 2.3

The symmetry center of the set of points of the conic (), is called the center of the conic ().

If the conic () would be written analytically as the following equation: a11x2 + 2a12xy + a22y2+ k = 0, then the origin of the frame would be the symmetry center of the conic, f(x,y) = f (-x,-y). Let us determine de conditions where the conic () admits a center, if so, well also try to find it.

Using the translation from frame R (O,) to the point C(xo,yo):

(2.4)

the equation (2.1) will be written under the following form:

a11x2 + 2a12xy + a22y2 + 2(a11xo + a12yo + a13)x +

+ 2(a21xo + a22yo + a23 )y + f(xo,yo) = 0

Applying the symmetry condition f(x,y) = f(-x,-y) we get:

or (2.5)

The equations (2.5) represent the equations of the center, if it exists.

The following cases are to be considered:

a) , system (2.5) has a unique solution, the point C(xo,yo) is the center of the conic ().

b) , system (2.5) has no solution, or admits an infinity of solutions, which means that the conic () has no unique center given a finite distance.

In order to recognize what the equation (2.1) geometrically represents, we will determine the frame in which the equation has its simplest form using an izometric transformation. This form will be called the canonical form of the conic (). This way, we have proven that the equation (2.1) is equivalent to one of the reduced equations described above, or with the empty set. The problem will be handled depending on the value of d which is either d 0 or d = 0 .



2.1 Reducing to the canonical form of conics with center , d 0

Consider a conic () analytically represented by the equation (2.1) with the center in point C(xo,yo). Translating the frame R (O,) to the point C(xo,yo) , (equations (2.4)) we obtain:

a11x2 + 2a12xy + a22y2 + k = 0 , k = f(xo,yo) (2.6)

Lets consider the quadratic form a11x2 + 2a12xy + a22y2 and the assosicated transformation T : E2 E2 having the same matrix as the quadratic form j. We know that an orthonormed frame exists, consisting of the eigenvectors of the transformation T, hence we can write the quadratic from j as a sum of squares :

l1X2 + l2 Y2, (2.7)

where l1 and l2 are the eigenvalues of the transformation T , solutions of the characteristic equation:

or l2 I l + d = 0 (2.8)

Equation (2.8) is called secular equation, and it always has a solution because the matrix (aij), i,j=1,2 is symmetrical.

Both frames are orthonormed, therefore the translation from

R (C,) with axes Cx and Cy to the frame R (C,) with axes CX and CY is obtained using an izometrical transformation trough C as a fixed point.

Noting with (x1,x2) and (h1,h2) the coordinates of the eigenvectors and in the frame R (C,) , the coordinate transformation is given by

, (2.9)

where the translation matrix is R = and is orthogonal, and has det.R = 1, meaning that the traslation from R to R is done by using a curl when det.R = 1, followed by a symmetry if det.R = - 1 .

In the new frame determined by the eigenvectors and , having the center of the conic as an origin, the equation of the conic (2.1) becomes:

l1X2 + l2 Y2 + k = 0 (2.10)

If we use the invariance of D at orthogonal transformations, and we calculate it for the carthesian frame XCY, we obtain D = k l1l2 = k d ( d =l1l2 from the secular equation) therefore, k = . Hence, in the frame R (C,) the equation of the conic () is written as:

(): l1X2 + l2 Y2 + = 0 (2.11)

and is called the canonical form of the equation of the conic.

Remarks:

1.      In the case of conics with center, the canonical equation (2.11) can be written easily if the orthogonal invariants D,d and I are known.

2. When reducing to the canonical from the equation of the conic with center, the order of the transformations is not essential: first the translation to the center of the conic and then the fixed point isometry, or viceversa.

3. The coordinate axes CX and CY are symmetry axes for the conic (g), therefore the frame R (C,) is the frame where the conic can be written under a reduced form, hence it can be recognized geometrically.

The slopes of these symmetry axes can be determined using the coordinate expression of the eigenvectors. If the eigenvectors and are normed, corresponding to the eigenvalues l1 and l2 then the frame

R (C,) is obtained through an angle rotation where aI [0,2p], from frame R (O,) if and only if for = (x1,x2) we have . Hence, we can write the equations which determine these eigenvectors:

and

noting m = tg a = we obtain

, which is equivalent to

a11 m2 + (a11 a22 ) m a12 = 0 , (2.12)

called the slope equation of the symmetry centers of a non-degenerated conic with the center at a finite distance. The equations of the axes of the conic, can be written as the line equations which pass through point C(xo,yo) and have the slopes m1 and m2 , the solutions of equation (2.12) .

4.                 To find the canonical frame using a translation followed by a rotation (curl) we will choose the order (and signs) if the eigenvectors accordingly, so that det.R = 1.

In order to determine the rotation angle, when passing from frame R (C,) to frame R (C,), we will write the equations which determine the eigenvectors:

and . (2.13)

Denoting the angle between with q I (0,) , we choose the sign of so that the agle between and to be q, hence

and

From relations (2.12) we get :

and , therefore

, hence

tg 2q = (2.14)

The same relations show that : , hence

sign(l1-l2) = sign (a12) .

5. By considering the analytical characteristics of the izometrical transformations, the reduction to the canonical form of a conic can be done using the roto-translation method. We operate the rotation

(2.15)

and we determine the angle j , where the coefficients of xy must be zero. After fulfilling this conditions and grouping the terms as a sum of squares, we can do the translation to the center of the conic.

6.      If we wish to draw the conic (g) the following steps have to be made:

- we write the conic (g) in the canonical form (2.11)

- we plot the center C(xo,yo) with respect to the xOy carthesian reference frame, then the center and the axes of the canonical reference frame CX and CY (with direction and sign), determined from the eigenvectors and plotted in point C.

- in the carthesian frame XCY we draw the conic (g) according to the canonical form (2.11)

Returning to the canonical form (2.11), lets analyze the following situations:

Case 1o D 0

a) d > 0 T l1l2 = d > 0 , equation (2.11) can be written in one of the following forms:

or , (2.16)

hence, we have the case of a real elypse, or the case of the empty set.

b) d < 0 T l1l2 = d < 0 , equation (2.11) can be written in one of the following forms:

or , (2.17)

hence the conic (g) represents a hyperbola. If I = a11 + a22 = 0 l1 = -l2 ,

that would mean a = b and the hyperbola (2.16) is equilateral.

Case 2o D = 0

a) d > 0 T l1l2 = d > 0 , the equation is written under the form

a2 X2 + b2Y2 = 0 . (2.18)

In such case, the conic is reduced to a point, the center C(xo,yo) .

b) d < 0 T l1l2 = d < 0 , equation (2.11) can be written in the form:

a2 X2 - b2Y2 = 0 (a X - bY) (a X + bY) = 0 , (2.19)

hence the conic is in fact two non-parallel lines.

As a conclusion, we can state that the cubic invariant D gives us information about the nature of the conic, and the square invariant d gives us information about the type of the conic (g). Therefore we can say:

if D 0 - we have a non-degenerated conic

D = 0 - we have a degenerated conic

and for

d > 0 - conic (g) is an elypse

d < 0 - conic (g) is a hyperbola

2.2 Reducing a conic without a (unique) center to its canonical form.

(d = 0)

Remember, that in case of d = 0 the system (2.5) is incompatible, or it admits an infinity of solutions, hence the conic (2.1) does not admit a symmetry point (the conic admits a line of centers). In such a case there is no translation that could lead us to an equation of second order, without first order members. Therefore, we will apply first an isometrical transformation with respect to the origin O as a fixed point (preferably a rotation) afterwards we apply a tranlation, which could suit our needs.

For d = 0 , the secular equation is written under the form l2 - Il = 0 with the solutions l1 = 0 and l2 = I . If the versors and , are eigenvectors which corespond to this transformation, then in the new frame R (O,), using the coordinate change:




(2.20)

the equation (2.1) will be written as:

l2 y2 + 2 a13x + 2a23y + a33 = 0 , l2 = I . (2.21)

Because the upper transformation is orthogonal, D is invariant with value D = - l2 (a13)2 , given by the associated matrix of the form (2.20) , therefore we obtain:

a13 = = (2.22)

Consider the following cases:

Case 1o D 0 T a13 0 , we apply the translation:

, (2.23)

choosing the point (xo,yo) so that the equation of the conic has the simplest form. Hence:

l2Y2 + 2a13X + 2(l2yo + a23)Y +

+ l2yo2 + 2a13 xo + 2 a23 yo + a33 = 0 (2.24)

We determine (xo,yo) with aid of these conditions:

(2.25)

The system (2.25) has a unique solution, and the equation (2.24) is written as:

Y2 =2pX , p = (2.26)

The conic (2.26), with respect to the frame R(V, ) , is a parabola with the edge in point V(xo,yo), the symmetry axis VX, and the tangent to its edge the axis VY .

Case 2o D = 0 T a13 = 0, equation (2.21) is written as:

l2 y2 + 2a23y + a33 = 0 , l2 = I . (2.27)

Equation (2.27) represents a polinom of second degree with variable y and the solutions k1 , k2 , either real or complex.

a) if k1, k2 I R , for k1k2 , the canonical form of the equation (2.27) is:

.

Applying the translation and noting the free term with - k2 , the conic has the following canonical form in the new frame:

Y2 - k2 = 0 , (2.28)

practically, 2 strictly parallel lines.

For k1 = k2 , after applying the translation we obtain

Y2 = 0, (2.29)

practically 2 parallel lines.

b) if k1, k2 R , equation (2.27) is equivalent to the empty set, therefore the conic (g) is represented in the plane as the empty set.

Those conics which have d= 0 are of the parabola type.

Remarks : 1. If d = a11a22 (a12 )2 = 0 , the second order members are a perfect square, and the equation (2.1) is a parabola.

2. When plotting a non-degenerated parabola, we must follow the following steps:

-determine the eigenvalues l1=0, l2= a11 + a22 and the corresponding eigenvectors ,

- we apply the rotation:

- the squares will be formed, and apply the translation to the point V(xo,yo)

- we draw the conic given by the canonical form Y2 =2pX in the frame having the origin on the edge of the parabola and the axes with direction and sign determined by the eigenvectors and .

For a degenerated parabola, we use the substitution

t = , (a11 a22 > 0 ) and we find the solutions t1 , t2 and we plot in the frame xOy , the two (real) lines.

3. If is the eigenvector corresponding to the eigenvalue l1 = 0, we have

, (2.30)

and by noting the angle between and , with q we obtain

tg q = , (2.31)

the formula which allows us to find the slope of the axis of the parabola.

As a conclusion, we can use the orthogonal invariants of a conic to write the the following isometrical classification:

D (nature)

d (type)

Comment

D 0

non-degenerated conics

d> 0

Real elypse, for I D < 0

Empty set, for I D < 0

d= 0

Parabola

d< 0

Hiperbola

D = 0

degenerated conics

d> 0

Double point

d= 0

Pair of lines, (parallel or stacked)

or the empty set.

d< 0

Pair of concurrent lines

2.3    Intersection between a line and a conic.

Consider the conic:

(g) : a11x2 + 2a12xy + a22y2 + 2a13x + 2a23y + a33 = 0 , a112+ a122+ a2220,

and the line which passes in the point M(xo,yo) and the direction given by = (l,h)

(d) : x = xo + lt , y = yo = ht , tI R .

In order to study thre relative position of the line (d) with respect to the conic (g) we must study the set of solutions of the system formed by the equations of the line, and equation of the conic. If we enter the coordinates of a point belonging to the line into the equation of the conic, we obtain the equation of second order in the variable t I R :

j(l,h) t2 + 2 ( l + h) t + = 0 (2.32)

where:

(2.33)

It is clear that a line intersects a conic in maximum two points.

We have the following cases:

Case 1o If j(l,h) 0 , for

a) ( l + h)2 - j(l,h) > 0, the equation (2.32) has two real solutions t1 t2 and the line interesects the conic in 2 points M1M2.

b) ( l + h)2 - j(l,h) = 0, the equation (2.32) has two identic real solutions t1=t2, and the line interesects the conic in two identic points M1=M2, hence the line is tangent to the conic in point M1.

c) ( l + h)2 - j(l,h) < 0, the equation (2.32) has no real solutions, therefore the line (d) does not interesect the conic (g) .

Case 20 if j(l,h) = 0, and

a) ( l + h) 0, the equation (2.32) has a unique solution and the line intersects the conic in a single point.

b) ( l + h) = 0 and 0 , the equation (2.23) has no solutions, therefore the line does not intersect the conic.

c) ( l + h) = 0 and=0 , the equation (2.23) is identically satisfied, hence the points of the line belong to the conic, situation possible if we have a degenerated conic (two lines)

Definition.2.4

The direction with the property

(2.35)

is called asymptotic direction of the conic (g)

Resulting from case 2o, a line with an asymptotic direction can intersect the conic in one single point at most.



Definition.2.5

A line that does not intersect the conic, and has asymptotical direction is called the asymptote of the non-degenerated conic (g).

Realizantul of the equation (2.35), with the variable m = , is given by (a12)2- a11a22 = - d. Therefore, if

d > 0 , (g) is an elypse and has no asymptotical directions

d = 0 , (g) is a parabola and admits a double asymptotical direction which is the direction of the parabola axis.

d < 0 , (g) is a hyperbola, and admits 2 asymptotical directions, and the lines of these directions pass through the center of the conic, and are the aymptotes of the hyperbola given by this equation:

a22 m2 + 2a12 m + a11 = 0 (2.36)

2.4    The Diameter of conic conjugated with a given line.

Polar and pole

Consider the conic (g) given by the equation (2.1) and a fixed direction given by the vector . The lines of the line family parallel to the direction (l,h), intersect the conic (g) in maximum 2 points. If M1 and M2 are the intersection points of the conic with a line of direction , then we have the following theorem:

Theorem 2.6

The geometrical locus of the middles of the segment M1M2 given by the secant lines of the conic (g ), with the direction are a subset of a line.

Proof. Consider the line (d) passing trough point Mo(xo,yo), with non-asymptotic direction

(d) : ,

intersecting the conic in the points M1(t1) and M2(t2). The values t1 and t2 corresponding to the intersection points M1 and M2 are the solutions of the equation (2.23). Without endangering the generality, let us consider the point MoId the middle of the segment M1M2 . The coordinates if the point Mo are

and , therefore,

t1 + t2 = 0 , and .

If Mo is the middle of the arbitrary cord of direction , the coordinates of the middles of the segment M1M2 satisfy the equation:

(2.37)

or

, (2.37)

practically, a line. Q.e.d.

Definition.2.7

The line conjugate to the non-asymptotic direction and given by the equation

(2.38)

is called the diameter of the conic (g).

If d 0 and arbitrary, the equation (2.38) represents the pencil of planes passing through the center of the conic (g), therefore the conjugate of a given direction is a line passing through the center of the conic.

If d = 0 and D 0, the equation (2.38) has the form fx+ l = 0, lIR, which represents a pencil of parallel planes with the direction , which is exactly the symmetry axis of the parabola. Therefore, the axis of the parabola is a conjugate of the diameter with the perpendicular direction to the axis, .

The diameter conjugate to the direction m= , equation (2.38) can be written in the following form:

(2.39)

and has the slope m given by

, or (2.40)

(2.40)

called relatia de conjugare .

Two lines with non-asymptotical directions, passing through the center of the conic (g) , (d0) and with their slopes m and m satisfy the equation (2.40) and are called the conjugate diametre of eachother.

Remarks:

1. The othogonal conjugate diametres define the symmetry axes of a conic with a center. Respecting the condition mm = - 1, from equation (2.40), we obtain the equations of the slopes of the symmetry axes of the conic (g)

,

hence, the same with equation (2.12) .

2. Self-conjugate diametres, m = m, define the asymptotes of the conic (g) From the equation (2.40) we obtain the equations of the slopes of the conic asymptotes (hyperbola)

, the same with equation (2.36) .

Let us consider the fixed point Mo(xo,yo) and (d), a line with variable direction passing through point Mo: x=xo+lt, y=yo+ht.

Consider now the points Mo , M1, M2 and M on the line (d) , characterized by the parametrical coordinates to, t1, t2 and t .

Definition.2.8

The point M is called the armonical conjugate of the point Mo with respect to M1 and M2 if

, (2.41)

MiMj being an oriented segment.

Relation (2.41) is written in parametrical coordinates like this:

or (2.42)

If the points M1 and M2 are the intersection points of the line (d) with the conic(g) , then the following theorem is true:

Theorem 2.9

The geometrical locus of point M , the armonical conjugate of point Mo with respect to points M1 and M2 , is a subset of the line with the following equation:

,

(2.43)

Proof. Using the relations (2.42) and (2.32) we obtain

(2.44)

By eliminating the parameters t, l , h from the equation (2.44) and the equations of the line (d) we obtain:

, hence well get equation (2.43)

Equation (2.43) can be obtained from the general equation of a conic (2.1) by using the following substitutions, called halvings:

si

Definition.2.10

The line with the equation (2.43) is called the polar of the point Mo with respect to the conic (g)

The point Mo whos polar with respect to the conic (g) is the line (d), is called the pole of the line (d).

Observation 3.

If point Mo(xo,yo) belongs to the conic (g) , then f(xo,yo) = 0 considering that the direction of the tangent through point Mo is given by , the equation is equivalent to the equation (2.43) , hence, the equation of the tangent in the point Mo .

2.5    Conics defined by initial conditions.

Consider the conic (g) given by the general equation (2.1)

f (x,y) s a11x2 + 2a12xy + a22y2 + 2a13x + 2a23y + a33 = 0, a112+ a122+ a2220.

The six coefficients from the equation of the conic cant be all zero at the same time, therefore this equation is equivalent with an equation which depends on only five coefficients, called esential parameters, obtained when dividing the initial six with one of them, non-zero. Therefore, in order to determine a conic uniquely, we need five conditions. For example a conic given by five points has the following equation:

(2.45)

Consider the conics : (g1) : f(x,y) = 0 and (g2) : g(x,y) = 0 .

Definition.2.11

The set of conics given by the general equation

(G ) , (2.46)

is called pencil of conics given by the fundamental conics (g1) and (g2) .

The conics (g1) and (g2) belong to the pencil (G ) but equation (2.46) does not always represent a conic (case of simple points and lines). Knowing that a and b are not zero at the same time we can write the equation (2.46) in the following form:

(G ) , mIR ,

pencil, which does not contain the conic (g2) .

If (g1) and (g2) intersect, then their intersection cannot contain more then four points. From earlier propsitions we conclude that through four non-collinear points pass an infinity of conics. If the distinct conics (g1) and (g2) are degenerate conics, degenerated into two lines, then the intersection points define in the plane a rectangle or a triangle.

Denoting with (AB) the left member of the equation of the line through A and B we have the following results:

a) The conics family circumscribed to the rectangle ABCD is given by the equation

a (AB) (CD) + b (BC) (AD) = 0 (2.47)

Particularly, the general equations of the conics which pass through the intersection of a conic f(x,y) = 0 with two lines D1 = 0 , D2 = 0, is given by:

a f(x,y) + b D1D2 = 0 (2.48)

If the line D = 0 intersects the conic f(x,y) = 0 in two lines, we can consired it as the intersection between the degenerated conic D2 = 0 and the conic f(x,y) = 0 , case in which the conics of the pencil (2.47) become bitangent to the conic f(x,y) = 0 , in its intersection points with the line D=0 therefore, we have:

a f(x,y) + b D2 = 0 (2.48)

In particular, the equation of the tangent conic to the lines D1 = 0 , D2 = 0 in the points where these lines are intersected by the line D = 0 , hence they form a triangle, is given by:

a D1D2 + b D2 = 0 (2.49)

b) Consider the conics (g1) : f(x,y) = 0 , (g2) : g(x,y) = 0 and

(g3) : h(x,y) = 0, then the set of conics given by the equation

a f(x,y) + b g(x,y) +m h(x,y) = 0

is called the family of conics given by the conics (g1) , (g2) and (g3) .

Therefore the pencil of conics circumscribed to the triangle ABC is given by:

a (AB)(AC)+ b(BA)(BC) +m(CA)(CB) = 0 . (2.50)








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