Locus of centers of conics through four points by geometry

Question

I found the solution to one of the interesting problems of conic sections a little while ago and then I searched through this site and found an analytical approach, I am now going to provide an engineering approach and I hope to get proof in the comments

We know that for five points in the plane, four of which do not lie on the same line, there is a single conic segment passing through these points, so there is an infinite number of conic sections that Passing through four points, the centers of this family of conic sections make a one-dimensional geometric shop that turns out to be a conical section. What I did was create five points of this conic cutting using a ruler and compass.

We are looking for the geometric locale of the center of the conical section that passes through the points: $A,B,C,D$ The desired geometric locus is the conical section that passes through the following five points:

$P=AB∩CD$

$Q=AD∩BC$

$K=AC∩BD$

$M$ is mid point $B,D$

$N$ is mid point $A,C$

Answer 1

OP claims a solution has already been obtained by using a coordinate system, but it is not presented. For clarity, I will derive this here, and show that the midpoints are always on the conic. For simplicity, we use translation invariance to set the origin at point D. The most general conic through the origin with center $(x_0,y_0)$ is then given by

$$a(x-x_0)^2+b(x-x_0)(y-y_0)+c(y-y_0)^2=ax_0^2+bx_0y_0+cy_0^2$$

The parameters are determined such that the conic passes through $A(x_1,y_1), B(x_2,y_2), C(x_3,y_3)$. Conveniently, the resulting system in $a,b,c$ is homogeneous:

$$M(x_0,y_0)\begin{pmatrix}a\\b\\c\end{pmatrix}=0$$

where $M$ is the $3\times 3$ matrix

$$M(x_0,y_0)=\begin{pmatrix}x_1^2-2x_1x_0&x_1y_1-x_0y_1-y_0x_1&y_1^2-2y_1y_0\\x_2^2-2x_2x_0&x_2y_2-x_0y_2-y_0x_2&y_2^2-2y_2y_0\\x_3^2-2x_3x_0&x_3y_3-x_0y_3-y_0x_3&y_3^2-2y_3y_0\end{pmatrix}$$ and the condition for the existence of a solution $\det M(x_0,y_0)=0$ yields the locus of the center. Somewhat surprisingly, this curve contains no cubic terms. This can be shown for example by splitting the matrix in 3 parts:

$$M=P+x_0 Q+ y_0 R$$

Then one can show using the Levi-Civita definition of the determinant for example, that since $Q,R$ have a different zero column, the $x_0^3, y_0^3, x_0^2y_0, x_0y_0^2$ terms must vanish identically. Terms of order $<3$ do not necessarily vanish, and the resulting locus is thus a conic section. To restore the coordinates of $D$, all one has to do is replace $x_i\to x_i-x_4, y_i\to y_i-y_4~~, i=1,2,3$.

The centers of $AC, AD, BD, BC, AB, CD$ always belong to the locus, as direct substitution of the coordinates of the midpoints of these yields vanishing results when substituted into the locus equation. The algebraic reason why they always belong to the locus is not particularly deep; when $(x_0, y_0)=(x_i/2, y_i/2)$, the i-th row of $M$ becomes identically zero and when $(x_0, y_0)=((x_i+x_j)/2, (y_i+y_j)/2)$ the i-th and j-th rows of the matrix become linearly dependent.

Below is some Mathematica code that yields the requisite locus:

vec = {x^2 + 
    2 x*Subscript[u, 0], (x*y - Subscript[u, 0]*y - 
     Subscript[v, 0]*x), y^2 + 2 y*Subscript[v, 0]};
mat = {vec /. {x -> Subscript[x, 1], y -> Subscript[y, 1]}, vec /. {x -> Subscript[x, 2], y -> Subscript[y, 2]}, vec /. {x -> Subscript[x, 3], y -> Subscript[y, 3]}};
Total@Flatten[Table[FullSimplify[SeriesCoefficient[ Det[mat], {Subscript[u, 0], 0, j}, {Subscript[v, 0], 0, k}]] * Subscript[u, 0]^j*Subscript[v, 0]^k,` {j, 0, 3}, {k, 0, 3}]]