pencil of conics

Two conics (http://planetmath.org/TangentOfConicSection)

$U=\mathrm{\hspace{0.33em}0}\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}V=\mathrm{\hspace{0.33em}0}$ (1)

can intersect in four points, some of which may coincide or be “imaginary”.

The equation

$pU+qV=\mathrm{\hspace{0.33em}0},$ (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

$p{L}_1{L}_2+qV=\mathrm{\hspace{0.33em}0}.$ (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

$p{L}_1{L}_2+q{L}_3{L}_4=\mathrm{\hspace{0.33em}0}.$ (4)

For any pair  $(p,q)$  of values, one conic section (4) passes through the four points determined by the equation pairs

$$L_1=0\wedge L_3=0,L_1=0\wedge L_4=0,L_2=0\wedge L_3=0,L_2=0\wedge L_4=0.$$

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

$$(-1,\mathrm{\hspace{0.17em}0}),(1,\mathrm{\hspace{0.17em}0}),(0,\mathrm{\hspace{0.17em}1}),(0,\mathrm{\hspace{0.17em}2}),(2,\mathrm{\hspace{0.17em}2}).$$

We can take the lines

$$2x+y-2=0,x-y+1=0,2x-y+2=0,x+y-1=0$$

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

$p(2x+y-2)(x-y+1)=(2x-y+2)(x+y-1).$ (6)

The conics passes through  $(2,\mathrm{\hspace{0.17em}2})$, if we substitute  $x:=2$,  $y:=2$;  it follows that  $p=3$.  Using this value in (6) results the equation of the searched conics:

$2x^2-y^2-2xy+3y-2=\mathrm{\hspace{0.33em}0}$ (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