pencil of conics

pencil of conics

Two conics (

http://planetmath.org/TangentOfConicSection

)

U

=

0

and

V

=

0

(1)

can

intersect

in four points, some of which may coincide or be “

imaginary

”.

The equation

p

⁢

U

+

q

⁢

V

=

0

,

(2)

where

p

and

q

are freely chooseable

parametres

, not both 0, represents the

pencil

of all the conics which

pass through

the four

intersection

points of the conics (1); see

quadratic curves

.

The same pencil is gotten by replacing one of the conics (1) by two lines

L

1

=

0

and

L

2

=

0

,  such that the first line passes through two of the intersection points and the second line through the other two of those points; then the equation of the pencil reads

p

⁢

L

1

⁢

L

2

+

q

⁢

V

=

0

.

(3)

One can also replace similarly the other (

V

) of the conics (1) by two lines

L

3

=

0

and

L

4

=

0

; then the

pencil of conics

is

p

⁢

L

1

⁢

L

2

+

q

⁢

L

3

⁢

L

4

=

0

.

(4)

For any pair

(

p

,

q

)

of values, one

conic section

(4) passes through the four points determined by the equation pairs

L

1

=

0

∧

L

3

=

0

,

L

1

=

0

∧

L

4

=

0

,

L

2

=

0

∧

L

3

=

0

,

L

2

=

0

∧

L

4

=

0

.

The pencils given by the equations (2), (3) and (4) can be obtained also by fixing either of the parametres

p

and

q

for example to

-

1

, when e.g. the pencil (4) may be expressed by

p

⁢

L

1

⁢

L

2

=

L

3

⁢

L

4

.

(5)

Application.

Using (5), we can easily find the equation of a conics which passes through five given points; we may first form the equations of the sides

L

1

=

0

,

L

2

=

0

,

L

3

=

0

and

L

4

=

0

of the

quadrilateral

determined by four of the given points.  The equation of the searched conic is then (5), where the value of

p

is gotten by substituting the

coordinates

of the fifth point to (5) and by solving

p

.

Example.

Find the equation of the conic section passing through the points

(

-

1

,

0

)

,

(

1

,

0

)

,

(

0

,

1

)

,

(

0

,

2

)

,

(

2

,

2

)

.

We can take the lines

2

⁢

x

+

y

-

2

=

0

,

x

-

y

+

1

=

0

,

2

⁢

x

-

y

+

2

=

0

,

x

+

y

-

1

=

0

passing through pairs of the four first points.  The equation of the pencil of the conics passing through these points is thus of the form

p

⁢

(

2

⁢

x

+

y

-

2

)

⁢

(

x

-

y

+

1

)

=

(

2

⁢

x

-

y

+

2

)

⁢

(

x

+

y

-

1

)

.

(6)

The conics passes through

(

2

,

2

)

, if we substitute

x

:=

2

,

y

:=

2

;  it follows that

p

=

3

.
Using this value in (6) results the equation of the searched conics:

2

⁢

x

2

-

y

2

-

2

⁢

x

⁢

y

+

3

⁢

y

-

2

=

0

(7)

The coefficients

2

,

-

1

,

-

2

of the second degree terms let infer, that this curve is a

hyperbola

with axes not

parallel

to the coordinate axes (see quadratic curves (

http://planetmath.org/QuadraticCurves

)).

Title

pencil of conics

Canonical

name

PencilOfConics

Date of creation

2013-03-22 18:51:07

Last modified on

2013-03-22 18:51:07

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

21

Author

pahio (2872)

Entry type

Definition

Classification

msc 51N20

Classification

msc 51A99

Related topic

QuadraticCurves

Related topic

LineInThePlane

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