# pencil of conics

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

 $\displaystyle U\;=\;0\quad\mbox{and}\quad V\;=\;0$ (1)

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

The equation

 $\displaystyle pU+qV\;=\;0,$ (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

 $\displaystyle pL_{1}L_{2}+qV\;=\;0.$ (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

 $\displaystyle pL_{1}L_{2}+qL_{3}L_{4}\;=\;0.$ (4)

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

 $L_{1}=0\;\land\;L_{3}=0,\quad L_{1}=0\;\land\;L_{4}=0,\quad L_{2}=0\;\land\;L_% {3}=0,\quad L_{2}=0\;\land\,L_{4}=0.$

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

 $\displaystyle pL_{1}L_{2}\;=\;L_{3}L_{4}.$ (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

 $(-1,\,0),\quad(1,\,0),\quad(0,\,1),\quad(0,\,2),\quad(2,\,2).$

We can take the lines

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

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

 $\displaystyle p(2x+y-2)(x-y+1)\;=\;(2x-y+2)(x+y-1).$ (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:

 $\displaystyle 2x^{2}-y^{2}-2xy+3y-2\;=\;0$ (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 PencilOfConics 2013-03-22 18:51:07 2013-03-22 18:51:07 pahio (2872) pahio (2872) 21 pahio (2872) Definition msc 51N20 msc 51A99 QuadraticCurves LineInThePlane