[alogo] Conic passing through the origin

Here I am interested in a conic with vanishing constant term (c=0 in Conic_Equation.html ):
(1)                                                       f(x,y) = ax2+2hxy+by2+2gx+2fy = 0.
This allows the conic to be written as a member of the family generated by pairs of lines. For example by factoring out x,wherever appears:
(2)                                                                 m(x,y) = x*(ax + 2hy + 2g),
and doing the same for the remaining terms involving y:
(3)                                                                     n(x,y) = y*(by +2f).
The family to which f(x,y)=0 belongs as a member is then determined by combining these products:
(4)                                                                     m(x,y) + r*n(x,y) = 0,    with arbitrary r.
The first degenerate conic m(x,y)=0 is the product of two lines: the y-axis (x=0) and the line ax+2hy+2g = 0 intersecting the y-axis at y = -g/h. The second degenerate conic n(x,y)=0 consists of two parallel lines: the x-axis (y=0) and the parallel to it  by+2f = 0 or y = -2f/b. The given conic is determined by taking r=1, i.e. finding a point (x,y) for which  m(x,y), n(x,y) have opposite values. This seems to be easier than solving the equation but is in fact equivalent to it. The trick works when there is an obvious point (x0, y0) satisfying the equation and being different from the four intersection points of the lines.

[0_0] [0_1] [0_2] [0_3]
[1_0] [1_1] [1_2] [1_3]
[2_0] [2_1] [2_2] [2_3]

The figure shows various members of the family but not a hint as to where to find a point lying on the given conic i.e.
satisfying f(x,y)=0. Anyway the analysis shows that the determination of the conic reduces by this procedure to finding a single point (x0, y0) satisfying the equation of the conic. The situation is not as bad as it looks however. In fact, consider a line through the origin  y = kx and try to find a second intersection  point of the given conic (different from O) on it.
Replacing this into the equation and simplifying we get:
             f(x,kx) = ax2+2hkx2+bk2x2+2gx+2fkx = 0 => ax + 2hkx + bk2x + 2g + 2fk = 0, which implies:
(5)                                                            x = -[2g+2fk] / [bk2 + 2hk +a].
This is nothing else than the rational parameterization of the conic which delivers the wanted fifth point (x,kx) .
The only restriction is to choose k to be different from the roots of the denominator, which are precisely (see Conic_Equation.html ) the directions of the asymptotes of the conic. Thus, selecting a convenient k we get easily at our fifth point. The exceptional cases for which either a=0 or b=0 are discussed in Conic_Passing_Origin2.html and Conic_Passing_Origin3.html .The generic example implementing the whole procedure drawing the concrete (generic) conic is contained in Conic_Instrument.html .

See Also



Return to Gallery

Produced with EucliDraw©