# 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:

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