f(t) = (at+b)/(ct

The set of points S={f(t)(

Inversely, every conic passing through the origin of coordinates and every set of two independent vectors {

To prove it start with the general equation of a conic in the system of coordinates defined by {

This equation is of the form g(x,y) = a

The condition to pass through the origin is equivalent with c

Take account of this and introduce the notation.

g(

Here

Assume now that f(t) is a function such that the point f(t)(

Then it must satisfy g( f(t)(

This, expanding and simplifying, leads to the equation.

f(t)[ (

Solving this for f(t) we see that it has the stated form f(t) = (at+b)/(ct

The inverse is equally trivial. In fact, assume that.

x = t(at+b)/(ct

From this follows that t = x/y. Introducing this into the second we find that (x,y) satisfies.

cx

This is the general equation of a conic passing through the origin of coordinates.

f(t) = (at+b)/(a

The conic passes through the origin O for the value t

The conic intersects the

A third point of the conic is easily determined from the parameterization. This is the point at which the conic crosses the

From the geometry it is easily seen that vector

Easily also is calculated the direction of the tangent for t=0 and is shown to be parallel to

The conic is then easily constructed as a member of the bitangent family of conics generated by line-pair (DO, DC) and the double line (OC). The member is completely determined through the requirement to pass through B.

Depending on the roots of the denominator we obtain ellipses, parabolas or hyperbolas. The corresponding images are to be found in ConicCharacterizationEllipse.html and ConicCharacterizationHyperbola.html .

ConicCharacterizationHyperbola.html

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