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.
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.