6 conics through 3 points and tangent to a line

Question

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!

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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 $x_1,x_2,x_3$.

Answer 1

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.

Answer 2

This answer explains why the family of conics passing through the points $A_1,A_2,A_3$ and tangent to a line L constitute a conic in $P$. It uses quadratic forms and lends itself to explicit computations. (I used a combination of Mathematica and Geogebra to develop and visualize this proof).

In what follows, we will discuss various families of conics incident with a given set of points and tangents. We'll use the shorthand $mP/nT$ to designate a set of $m$ points and $n$ tangents.

1. Background

The conic $C$ defined by the equation $ ax^2+cy^2+2bxy+2dx+2ey+f=0 $ can be expressed as a quadratic form $p^TAp=0,$ where point $p=\begin{bmatrix} x \\ y \\ 1 \\ \end{bmatrix} $ and $A= \begin{bmatrix} a & b & d \\ b & c & e \\ d & e & f \\ \end{bmatrix} $ is a symmetric matrix.

For a point $p$, $Ap$ is the polar of $p$ with respect to $C$ (See Wikipedia: Pole and Polar). If $p$ is on $C$, then $Ap$ is the tangent to $C$ at $p$.

Dualizing from points to lines, the tangential equation of the same conic can be written as $l^TA^{-1}l=0$, where $l=\begin{bmatrix} r \\ s \\ t \\ \end{bmatrix} $ represents the line $rx+sy+t=0.$

The matrix $A^{-1}$ has a messy denominator and can be replaced by the simpler adjoint (aka adjugate) matrix $A^\triangle=A^{-1}\operatorname{det}(A).$

The following excerpt from Richter-Gebert, Perspectives on Projective Geometry, pg 155 gives more detail and saves me some typing:

If the above reference is not available, see Wikipedia for more about inverses, adjoints, cofactors, etc.

2. The Locus of a 3P/1T Family of Conics

The $3P$ family of conics incident with three points $A_1,A_2,A_3$ form a projective plane $P$, each point of which is a conic. Given three conics $C_1,C_2,C_3$ passing through $A_1,A_2,A_3$, a conic $C$ in $P$ can be written as $C_1+aC_2+bC_3$ (technically we should use a homogeneous sum $aC_1+bC_2+cC_3$ but the argument remains the same).

If we write $C$ as a quadratic form $p^TAp,$ the entries of $A$ are linear polynomials in $a,b$ (equivalently, linear combinations of $1,a,b$). Given a line $l=\begin{bmatrix} r \\ s \\ t \\ \end{bmatrix}$, a conic in $P$ that is tangent to $l$ will satisfy the equation $l^TA^{\triangle}l=0.$

The elements of $A^{\triangle}$ are quadratic polynomials in $a,b$. For example (see section 1), the top left element $a$ of $A^{\triangle}$ is the determinant $ \begin{vmatrix} c & e \\ e & f \\ \end{vmatrix}, $ where $c,e,f$ are elements of $A$. Since the latter are linear polynomials, $a$ will be a quadratic polynomial. And thus the expression $l^TA^{\triangle}l$ will evaluate to a quadratic polynomial in $a,b$. Recalling that $a,b$ are coordinates for $P$, we see that the locus of conics incident with $A_1,A_2,A_3$ and tangent to $l$ is itself a conic in $P$.

3. Remarks

I was initially surprised by the OP's observation that the family of conics in question defined a conic in the projective plane of conics passing through three points. The literature on projective geometry and conics frequently talks about pencils and conics that pass through four base points. These pencils are straight lines in $P^5,$ the five dimensional projective space of all conics. But I have never seen a description of the locus of conics incident to three points and a line.

The reference Pamfilos, A Gallery of Conics by Five Elements reviews conics defined by five elements, i.e., either lines to which the conic is tangent or points through which the conic pass. Pencils (and their dual: ranges) are discussed in section 1.2. Conics defined by four points and a tangent (4P/1T) are discussed on pg 304. As seen in Figure 1.2, there are always two conics which fit the bill.

I would speculate that the reason that the 4P/1T has two results is related to the fact that the locus for 3P/1T is a degree 2 curve. The same way that the existence of a unique conic for the case 5P is related to the fact that the locus for 4P is a straight line.