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# pencil of conics

Two conics

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

## Mathematics Subject Classification

51N20*no label found*51A99

*no label found*

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