[alogo] Conics tangent and intersecting

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.


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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 :


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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.
This shows that the intersection point H = B+rC on BC is independent of the particular member ct of the conic family hence the same for all.
Remark The figure displays also the triangle conic cK with perspector K belonging to the considered family of conics. It is obtained for t=1. A sequel to this is the discussion on a related nodal cubic in CubicFromTwoConics.html .

See Also

ProjectiveBase.html
CubicFromTwoConics.html

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