I played in GeoGebra and discovered an interesting fact:
Let $L$ be a line.
Let $A_1,A_2,A_3$ be points that are not collinear and not on $L$.
Let $C_1,\dots,C_6$ be 6 conics that pass through points $A_1, A_2, A_3$ and are tangent to $L$.
Let $C_1,C_2$ intersect at a fourth point $Q_1$ other than $A_1,A_2,A_3$.
Let $C_2,C_3$ intersect at a fourth point $Q_2$ other than $A_1,A_2,A_3$.
Let $C_3,C_4$ intersect at a fourth point $Q_3$ other than $A_1,A_2,A_3$.
Let $C_4,C_5$ intersect at a fourth point $Q_4$ other than $A_1,A_2,A_3$.
Let $C_5,C_6$ intersect at a fourth point $Q_5$ other than $A_1,A_2,A_3$.
Let $C_6,C_1$ intersect at a fourth point $Q_6$ other than $A_1,A_2,A_3$.
Let $D_1$ be the conic through 5 points $Q_1, Q_4, A_1, A_2, A_3$.
Let $D_2$ be the conic through 5 points $Q_2, Q_5, A_1, A_2, A_3$.
Let $D_3$ be the conic through 5 points $Q_3, Q_6, A_1, A_2, A_3$.
Then $D_1, D_2, D_3$ pass through a fourth point $E$ other than $A_1, A_2, A_3$.
(The black conics are $C_1,\dots,C_6$. The red conics are $D_1,D_2,D_3$.)
This result, about 6 conics, is an analog of Pascal's theorem about 6 points.
I'd like to see a solution in any form. Thanks!
I was thinking that the conics through 3 points, form a projective plane $P$, so a conic is a point in $P$, and those conics additionally tangent to line $L$ is points on a conic in $P$, so this theorem is Pascal's theorem about 6 points on a conic in $P$.
I can give a proof of this but not sure if it is correct: A conic in $P$ is of the form$$B(r,s)=x_1C_1(r,s)+x_2C_2(r,s)+x_3C_3(r,s)$$ and $$\alpha u+\beta v$$ is a point on the line through $u,v$.
If the conic $B(r,s)=0$ is tangent to the line through $u,v$ then equation$$B(\alpha u+\beta v,\alpha u+\beta v)=0$$in $[\alpha:\beta]$ has a multiple root, then the determinant$$B(u,v)^2-B(u,u)B(v,v)=0$$Rewrite by the definition of $B(r,s)$: $$\small(x_1C_1(u,v)+x_2C_2(u,v)+x_3C_3(u,v))^2-(x_1C_1(u,u)+x_2C_2(u,u)+x_3C_3(u,u))(x_1C_1(v,v)+x_2C_2(v,v)+x_3C_3(v,v))$$This is a conic in $x_1,x_2,x_3$.
Best Answer
OP has observed that the set of conics passing through three points form a projective plane $P$, and has observed that a certain configuration of conics in $P$ give a result analogous to Pascal's theorem. OP also claims (1) that the family of conics passing through the points $A_1,A_2,A_3$ and tangent to a line L constitute a conic in $P$.
The purpose of this answer is to spell out how Pascal's theorem yields the result in question, assuming that claim (1) is true.
The figure above shows the projective plane $P$. A point of $P$ represents a conic passing through the points $A_1,A_2,A_3$. A line of $P$ represents a pencil $x$ of conics passing through base points $A_1,A_2,A_3,X$. Note that the name of the pencil and the fourth base point are the same, except for case.
The diagram follows the construction given in the OP, starting with 6 conics $C_i$ sitting on a conic in $P$, and ending with the (red) Pascal line $e$. For example, $q_1$ is the pencil of conics with base points $A_1,A_2,A_3,Q_1$, and $q_4$ is the pencil of conics with base points $A_1,A_2,A_3,Q_4$, and $D_1=q_1\cap q_4$ is the conic through the five points $Q_1, Q_4, A_1, A_2, A_3.$ Because $e$ represents a pencil of conics with base points $A_1,A_2,A_3,E$, we can conclude that $D_1, D_2, D_3$ pass through a fourth point $E$ other than $A_1, A_2, A_3,$ which was the result to be shown.