There are two approaches to the algebra. We can begin with P = p:q:r on TC
and try to compute everything else, or we can begin with the centre as U = u:v:w
on LI the line at infinity. Each has its advantages.

(A) P on TC.

Then instantly, Q = [1/p], cP = [q+r], cQ = [p(q+r)], P(P,Q) = [ p(q²-r²)].
We now proceed to show that R is [(q-r)/(q+r)], and that the centre K is [p(q-r)²].

For the moment write R' for the point [(q-r)/(q+r)].

We use two easy identities. If f(p,q,r) =-(q+r)(r-p)(p-q) or p(q-r)², then
f(p,q,r)+f(q,r,p)+f(r,p,q) = x²(y+z)y²(z+x)+z²(x+y)-6xyz, the Tucker cubic.

If we know that P is on TC, the second shows that K is on LI.
The first shows that R' is on SE and K is on P(P,Q).
Once we show that C(P,Q) is a parabola, K is the centre.

Another easy calculation shows that R' is the intersection of the polars of P,Q
with respect to SE. These polars are the duals of cP and cQ. As the intersection
of polars, R' is the pole of PQ

We now have from the first identity:

P on TC if and only if PQ is tangent to SE.

Now PQ is the isotome of C(P,Q) for any P since it contains the isotomes P and Q
which are on the conic. And the isotome of PQ is a parabola if and only if PQ is a
tangent to SE. Thus we have:

C(P,Q) is a parabola if and only if PQ is tangent to SE.

All of Tucker from one identity.

On the way, we saw that :

(1) C(P,Q) is the isotome of PQ.

From now on we will assume that P is on TC. Then R, the point of contact of PQ
with SE is indeed R'. We note that K = tR, so we can write K = [(q+r)/(q-r)] as
an alternative to the earlier version.

Knowing R, we can verify that :

(2) P(P,Q) = cR.

We now prove together

(3) C(P,R) and C(Q,R) are parabolas.
(12) P is the third Tucker point of C(P,R), Q is the third Tucker point of C(Q,R).

Proof
Let X = [(q-r)²] be the perspector of Circumconic C.
As X is the barycentric square of a point on LI, C is a parabola.
As Q is on TC, Q is on C. Clearly R is on C. Thus C = C(Q,R).
As P = tQ is on C(P,Q), it is not on C(Q,R).
Thus Q - on TC - is the third Tucker point of C(Q,R).
The result for C(P,R) is similar.

From the proof, we also have

(21) P(P,R) = [p²(q-r)²], P(Q,R) = [(q-r)²].

A quick calculation shows that P(P,R)P(Q,R) is the tripolar of R and so
as tR is the perspector of D, and G is on D,

(14) the line L = P(P,R)P(Q,R)
(a) touches D at G and is the isotome of D,
(b) is the tripolar of R.

Now R is on LI, so G is on L. Also P(P,R),P(Q,R) are on SI, so

(4) P(P,R) and P(Q,R) are antipodes on SI.

Now, complementation is an affine transformation - it fixes LI. Thus

(5) cPcQ is tangent to SI at P(P,Q) and is parallel to the isotome PQ.

Now we can combine (12), (22) - section (B) below - and (4) to get

(8) the tangents to C(P,R) at P,and C(Q,R) at Q are parallel.

Then using (13) - see section (B) - we get most of

(9) the tangents in (8) are also tangent to SE
their contact points with SE lie on L = P(P,R)P(Q,R).

The last bit follows by anticomplementation - the perspector maps to the
contact point with SE, as above. Then as G is on L, so is the contact point.

From the formulae in the proof of (3)(12), we have

(10) the tangents in (8) are the isotomes of C(P,R), C(Q,R).

Now let us consider the Circumconic D.
This has centre P(P,Q) = [p(q²-r²)] and perspector K = [p(q-r)²].
It follows that it has the infinite points [p(q-r)] and [(q-r)].
Its centre lies on the tripolars of P and Q. The tripolar of P also passes
through [p(q-r)], P(P,Q) and through P(P,R) as P is on TC.
The tripolar of Q passes through [q-r], P(Q,R) and through P(Q,R). Thus

(23) the following pairs of lines are identical
(a) the asymptotes of D,
(b) the tripolars of P and Q,
(c) the lines P(P,Q)P(P,R) and P(P,Q)P(Q,R).

Then using (25) from section (B),

(24)
the axis of C(P,R) is parallel to P(P,Q)P(P,R),
the axis of C(Q,R) is parallel to P(P,Q)P(Q,R).

(28) The tangents from P, Q to SE other than PQ are parallel.
The contact points are K = [1/(q-r)] and K' = [1/p(q-r)].
G,K,K',P(P,R) and P(Q,R) are collinear.

Proof
It is easiest to check that, as P, Q are on TC, they lie on the tangents
with the given contact points. We then observe that K' is on GK, so that
the tangents are parallel.
For the last bit, note that P(P,R) is the complement of K', P(Q,R) that of K.

(B) U on LI.

Then D, the Circumconic with perspector U meets C(P,Q) at S = [u/(v-w)].
It is easy to check that S lies on both conics.

The tangent to C(P,Q) at S is (v-w)²x+(w-u)²y+(u-v)²z = 0, and it is
easy to see that this is tangent to SE at T' = [1/(v-w)].
From (1), we know that PQ is the isotome of C(P,Q), so is u²x+v²y+w²z = 0.
As U is on LI, u+v+w = 0, and we see that [u(v-w)] is the infinite point on both lines.
We now have

(13) the tangent to C(P,Q) at S is parallel to PQ, and is tangent to SE at T'.

A simple Maple check shows that, as U is on LI, S is on TC. We have

(11) the third Tucker point of a circumparabola C is the fourth
intersection of C and D, where D has perspector the centre of C.

We then have, as PQ is the isotome of C(P,Q),

(24) the third Tucker point of a circumparabola C is the contact point
of the tangent parallel to the isotome of C.

(25) the third Tucker point of a Circumparabola is C the unique point
on C whose tripolar is parallel to the axis of C. The tripolar is PU,
where P is the perspector and U the centre of C.

Proof
The axis of a Circumparabola is parallel to any line through its centre.
We observe that U lies on the tripolar of S. The tripole of Z on C(P,Q) will
pass through U if and only if Z is on the Circumconic D (perspector U).
As we have seen, this is the point S.
The last is a simple calculation.

(27) the third Tucker point of a Circumparabola C is the second
intersection of C with GU, where U is the centre of C.

Proof
This is simple, the line GU is (v-w)x+(w-u)y+(u-v)z = 0.

In the Tucker Cubic, we discuss the notion of tripolar centroid.
This gives yet another characterisation.

(29) the third Tucker point of a Circumparabola C is the unique
point on C whose tripolar centroid is the perspector of C.

The proof is on the Cubic page, though it is easy to verify that S has the
required tripolar centroid. The uniqueness is less obvious.

Now let V = [(w-v)/u], the isotomic conjugate of S, so V is on PQ as S is
on C(P,Q). In this section, R = 1/u, and this has tripolar L - see above.
As S on TC, so is V. Thus,

(17) TC, PQ and L concur at V.

The tripolar of V is parallel to L, but appears to contain no useful points
except the antipode of P(P,Q) on SI. A calculation shows that the tripolar
of V meets GR at W = [(v-w)²]. This is on SI, as [v-w] is on LI.
Now P(P,Q) is on GR as the complement of R,

(26) the tripolar of V is parallel to L and meets P(P,Q)R at the antipode
of P(P,Q) on SI.

Of course, SV is the isotome of the Circumparabola C(S,V). A calculation
shows that SV has equation kx+ly+mz = 0, where [k] = [u²(v-w)²].
We recognize this as the tangent to SE at T, where T is the fourth intersection
of C(P,Q) and SE, so T = [1/u(v-w)].

We have proved

(15) SV is tangent to SE at T.

Note that T' = [1/(v-w)] is the fourth intersection of D and SE. We can check that

(16) TT' is tangent to D at T'.

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