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![[parent]](http://images.planetmath.org:8080/images/uparrow.png)
pencil of conics
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(Definition)
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Two conics
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(1) |
can intersect in four points, some of which may coincide or be ``imaginary''.
The equation
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(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
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(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
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(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
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(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
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(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:
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(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).
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"pencil of conics" is owned by pahio.
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Cross-references: parallel, hyperbola, curve, terms, degree, coefficients, coordinates, quadrilateral, sides, application, passes through, lines, quadratic curves, intersection, pass through, conics, pencil, represents, parametres, equation, points, intersect
This is version 17 of pencil of conics, born on 2009-03-08, modified 2009-03-25.
Object id is 11661, canonical name is PencilOfConics.
Accessed 641 times total.
Classification:
| AMS MSC: | 51A99 (Geometry :: Linear incidence geometry :: Miscellaneous) | | | 51N20 (Geometry :: Analytic and descriptive geometry :: Euclidean analytic geometry) |
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Pending Errata and Addenda
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