## 1. Conic characterization through a rational function

Given are two independent vectors {E, e} and a rational function of the form.
f(t) = (at+b)/(ct2+dt+e).
The set of points S={f(t)(E+te)} describes parametrically a conic passing through the origin of coordinates.
Inversely, every conic passing through the origin of coordinates and every set of two independent vectors {E, e} defines such a function f(t) so that the corresponding set S coincides with the original conic.

To prove it start with the general equation of a conic in the system of coordinates defined by {E, e}.
This equation is  of the form  g(x,y) = a1x2 + 2a2xy + a3y2 + b1x+b2y + c0 = 0.
The condition to pass through the origin is equivalent with c0 = 0.
Take account of this and introduce the notation.
g(x) = xtUx + Vtx      (*).
Here x=(x,y),  U is the matrix with rows (a1, a2) and (a2, a3) and V is the vector (b1, b2).
Assume now that f(t) is a function such that the point f(t)(E+te) is on the conic.

Then it must satisfy g( f(t)(E+te) ) = 0.
This, expanding and simplifying, leads to the equation.
f(t)[ (E+te)tU(E+te)] + Vt [E+te] = 0.
Solving this for f(t) we see that it has the stated form f(t) = (at+b)/(ct2+dt+e).

The inverse is equally trivial. In fact, assume that.
x = t(at+b)/(ct2+dt+e),     y = (at+b)/(ct2+dt+e).
From this follows that   t = x/y. Introducing this into the second we find that (x,y) satisfies.
cx2 + dxy + ey2 -ax - by = 0.
This is the general equation of a conic passing through the origin of coordinates.

## 2. Reconstructing the conic

To reconstruct the conic from the function f(t) I changed slightly the notation and write.
f(t) = (at+b)/(a1t2+b1t+c1).
The conic passes through the origin O for the value  t1 = -b/a.
The conic intersects the e-axis a second time when t goes to infinity. The point B at which the conic crosses the e-axis is  B = t2e, where t2 = a/a1.
A third point of the conic is easily determined from the parameterization. This is the point at which the conic crosses the E-axis. This occurs for t=0, and the point is C=t3E, with t3 = b/c1.
From the geometry it is easily seen that vector E+t1e defines the tangent at the origin.
Easily also is calculated the direction of the tangent for t=0 and is shown to be parallel to E+t4e, where t4 = (bc1)/(ac1-bb1). Let D be the intersection of the two tangents at O and C correspondingly.
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 .