We pick two lines and call them the axes, letting the point (a,0) on the first correspond with the point (0,(pa+q/ra+s)) on the second. Since we have not yet specified any basis points on these axes, we will do so in such a way that (infinity,0) corresponds to (0,0) and (0,0) corresponds to (0,infinity). Then the fractional linear transform a -> (pa+q)/(ra+s) will reduce to the form a -> k/a.
We want to find points on these lines such that their locus is
a curve tangent to all of the lines. On any line l(a), the point
we want is lim (l(a).l(a+r)) as r->0.
So, say we have two lines:
x y x y
- + --- = 1 and --- + ------- = 1
a k/a a+r k/(a+r)
After a crossmultiplication:
kx + (a+r)(a+r)y = ka + kr
kx + (a )(a )y = ka
After a subtraction: (2ar+rr)y = kr
Dividing by r: (2a+r)y = k
Thus y = k/(2a+r);
x = a - aay/k = a - aa/(2a+r).
We take
the limits of these quantities as r approaches 0, and find that
(x,y)=(a/2,k/2a). So xy-k/4=0. This equation states that a
particular quadratic function of x and y is zero; therefore
it represents a point conic.
Theorem: A Point Conic is a Line Conic
We now choose our axes in the affine plane so that our conic has the equation xx+yy=1. Then xdx + ydy = 0; dy/dx = -x/y. We now use the parameterization x=cos(t), y=sin(t); then our tangent line will be
y - sin(t) dy -cos(t) ---------- = -- = ------- = -cot(t) x - cos(t) dx sin(t)This intersects the tangent line x=-1 at y=(cos(t)+1)/sin(t)
Then x=(y+1)/(y-1), and the sets of points on the ranges are related by a fractional linear transformation. So they are also related by a projectivity, and indeed the line conic is a point conic.