K = cK(#F,R), with F = [f,g,h], R = [r,s,t].

If U is on C(F), then F is on T(U). Thus T(U) meets K twice at F, and at a third point U'.
T(U) is a nodal tangent if and only if U' = F.

Result 1 U' is the intersection of T(U) and T(U") where U" is the U²-isoconjugate of R.

Proof
We can check that U' = [u',v',w'], with u' = u(s/v-t/w), v' = v(t/w-r/u), w' = w(r/u-s/v)
satisfies the equations of the tripolars.
Now
   hv'-gw'
= hv(t/w-r/u)-gw(r/u-s/v)
= vw(ht/w²+gs/v²-(r/u)(h/w+g/v))
= vw(fr/u²+gs/v²+ht/w²), as U is on C(F).
Note that the third factor is symmetric in the variables.
Then it is easy to check that U' is on K.

Result 2
U' = F if and only if U is an intersection of C(F) and rf/x²+sg/y²+th/z² = 0.
There are two such points.

Proof
U' = F if and only if hv'-gw' = gu'-fv' = fw'-hu' = 0.
The result follows from the above proof.
The fact that there are two roots is best seen by looking at the equations in
(1/x,1/y,1/z), when we get the intersection of a line and conic.

The map U to U' maps U on C(F) to the third intersection of T(U) and K.
For X ≠ F on K, the inverse maps X to X^, the tripole of XF, since X is the third
intersection of the tripolar of X^ and K, and X^ is on C(F).

If X = [k,l,m], X^ has first barycentric 1/(gm-hl).

For any X, X* is the F²-isoconjugate of X. If X is on K, X* is on K.

Result 3
Points X, X* on K correspond to U, V on C(F) with R on UV.

Proof
X* corresponds to V with first barycentric f/k(gm-hl). Then UV has equation
k(gm-hl)²x+l(hk-fm)²y + m(fl-gk)²z = 0.
Thus R is on UV if and only if X is on K.

As a corollary, we see that we get a self-isoconjugate point - this must be F -
precisely when U is the contact point of a tangent from R to C(K). Then the
tripolar of U will be a nodal tangent.

cK0 conics