Consider two conics {c1,c2} having a common tangent at a point A and also intersecting at two points {B,C}. Then for every line through A intersecting the conics at {F1,F2} the tangents at these points intersect at a point H on line BC.
Let {P1,P2} be the intersection points of the tangents to conics {c1,c2} at their points {F1=A+p1E, F2=A+p2E} with line m. Points {F1,F2} lie on a variable line AE, E=B+kC. The intersection point G(k) of lines {F1P2,F2P1} describes a cubic passing through points {A,B,C,D}.
The discussion here is a sequel to the one in ConicsTangentIntersecting.html where the first part of the claim is proved. For the seek of convenience I repeat some facts from this discussion here:
Let D be the intersection point of the common tangent (m) with line BC. I use a projective base consisting of {A,B,C,K}, K being some point on the harmonic conjugate of line AD with respect to the pair of lines {AB,AC}. This simplifies some calculations done below.
In this system (see ProjectiveBase.html ) points {A,B,C,K} have correspondingly coordinates {(1,0,0),(0,1,0),(0,0,1),(1,1,1)}.
The intersection point D' of AK with BC has in this system the representation D' = B+C and D = B-C.
Then line m = AD is described by the equation : y + z = 0 and line m' = AD' by the equation y - z = 0.
Conics c1 and c2 belong to the family generated by the two degenerate conics consisting of pairs of lines:
c : yz = 0 (lines AC, AB) and c' : x(y+z) = 0 (lines BC, m).
The general conic ct of this family is determined by a parameter (t) in :
ct : yz + t x(y+z) = 0 => yz + txz + txy = 0.
Let E be a point on line BC : E = B + kC. The second intersection point F of ct with line AE : F = A + pE is determined by introducing F into the conic equation. Later, written as a quadratic form with the matrix M :
c(F,F) = 0 => c(A+pE,A+pE)=0 => 2pc(A,E) + p2c(E,E) = 0 => p = -2c(A,E)/c(E,E).
This replacing E = B+kC and making the matrix multiplications implies
p = -2c(A,B+kC)/c(B+kC,B+kC) = -c(A,B+kC)/(kc(B,C)) = -t(1+k)/k.
On the other side the tangent of the conic at F is given by equation c(F,X) = 0.
Its intersection point with line BC is found by replacing H=B+rC into last equation:
c(F,B+rC) = 0 => c(A+pE,B+rC) = 0 => c(A+p(B+kC),B+rC)=0 => c(A,B)+rc(A,C)+pkc(C,B)+rpc(B,C)=0.
Introducing the values from the matrix we obtain: t + rt + pr + pk = 0 => r = -(pk+t)/(p+t) = -k2.
Taking point E = B+kC on BC as before, the above calculations show that
{p1=-t1(1+k)/k, p2=-t2(1+k)/k}.
Point A in terms of F's (modulo multiplicative constant) is
A = p2 F1 - p1 F2. And the harmonic conjugate I of A with respect to {F1,F2}:
I = p2 F1 + p1 F2 = ... = (p1+p2)A + 2p1p2E.
Using the calculation of the previous section for H = B-k2C
the intersection point J of line HI with line m : y+z = 0 results by
applying y+z = 0 to the parametric expression for line
HI : (B -k2C)t + (p1+p2)A + 2p1p2(B+kC).
t -k2t + 2p1p2 + 2p1p2k = 0 => t = 2p1p2/(k-1).
Thus J = Ht + I => H = (1/t)(J - I) =>
G = (1/t)(J + I) = (1/t)(Ht + 2I) = H+(2/t)I = H + ((k-1)/(p1p2))((p1+p2)A + 2p1p2(B+kC)) =>
G(k) = [k(1-k)(t1+t2)/(t1t2)]A + [(2k-1)(1+k)]B + [k(k-2)(k+1)]C.
It is readily seen that for k one of {-1, 0, infinity, 1} the value of G is correspondingly {A,B,C,D}. Remark Note that the cubic depends only on the ratio s=t1/t2 of the parameters, hence is the same with the cubic resulting from t1=s and t2=1, the second corresponding to the conic with perspector K.
The tangent of the cubic at A is line m = AD. The tangents to the cubic at points {B,C,D} meet at a point of line AK. D is an inflexion point of the cubic.
The claim follows by computing the direction of the tangent at a point of the curve using the parameterization of the previous section.
In fact, the tangent at a point (x0,y0,z0) of a curve f(x,y,z)=0 is given by ux+vy+wz=0, with
These coefficients can be found from the parametric form of the curve G(k)=(Gx(k),Gy(k),Gz(k)) and the relations:
The first results from the Euler formula for homogeneous functions and the second is the chain rule. Thus, the coefficients of the tangent at G(k) are given by a vector product (Gx,Gy,Gz) x (G'x,G'y,G'z). Taking into account that
G'(k) = [(1-2k)(t1+t2)/(t1t2)]A + [4k+1]B + [3k2-2k-2]C
we find that (for brevity taking T=(t1+t2)/(t1t2)):
- Tangent at A (k=-1) : y+z = 0,
- Tangent at B (k=0) : 2x + Tz = 0,
- Tangent at C (k=infinity) : 2x + Ty = 0,
- Tangent at D (k=1) : 4x +Ty + Tz = 0.
First tangent is line m=AD. The other three intersect at the same point (-T, 2, 2) lying on AD' : y-z=0.
Applying the equation of the tangent at D : f(x,y,z)=4x+Ty+Tz at a point G(k) of the cubic we find f(G(k))=(k-1)3. It follows that near k=1 the curve has points on both sides of the tangent, hence D is a flex.
In fact, since the cubic is a rational one it is singular and has a node. To prove the claim apply the formula y-z=0 for line AD' to G(k) to obtain the equation (k+1)(k2-4k+1)=0, with roots:
The first solution corresponds to point A and the two other solutions define the same point (-T,3,3) on the curve. This is seen by a simple calculation.
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