(1) f(x,y) = ax

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 (x

The figure shows various members of the family but not a hint as to where to find a point lying on the

satisfying f(x,y)=0. Anyway the analysis shows that the determination of the conic reduces by this procedure to finding a

Replacing this into the equation and simplifying we get:

f(x,kx) = ax

(5) x = -[2g+2fk] / [bk

This is nothing else than the

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

Conic_Passing_Origin2.html

Conic_Passing_Origin3.html

Conic_Instrument.html

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