Question about conic sections and determinant Now I have found questions which were on the same topic on this site..but not quite one like this.
Let there be six points $p_1=(x_1,y_1),..,p_6= (x_6,y_6) $
Why is it, that if $$ det\begin{pmatrix}1 &1 &1 & 1& 1 & 1 \\ x_1 & x_2 & x_3 &x_4 & x_5 & x_6 \\
y_1 & y_2 & y_3 &y_4 & y_5 &y_6 \\ x_1^2 & x_2^2 & x_3^2 &x_4^2 & x_5^2 & x_6^2 \\ x_1y_1 & x_2y_2 & x_3y_3 & x_4y_4 & x_5y_5 & x_6y_6 \\ y_1^2 & y_2^2 & y_3^2 &y_4^2 & y_5^2 &y_6^2
\end{pmatrix}=0$$
All of those six points define a conic section?
I know that the matrix is derived from the general conic section formula $ax^2+bxy+cy^2+dx+ey+f=0$
and that a conic section is defined by five points, but here I stumble for an explaination.
 A: Let us consider the following homogeneous system with the matrix you have, but transposed :
$$\begin{pmatrix}1 &1 &1 & 1& 1 & 1 \\ x_1 & x_2 & x_3 &x_4 & x_5 & x_6 \\
y_1 & y_2 & y_3 &y_4 & y_5 &y_6 \\ x_1^2 & x_2^2 & x_3^2 &x_4^2 & x_5^2 & x_6^2 \\ x_1y_1 & x_2y_2 & x_3y_3 & x_4y_4 & x_5y_5 & x_6y_6 \\ y_1^2 & y_2^2 & y_3^2 &y_4^2 & y_5^2 &y_6^2
\end{pmatrix}^T\begin{pmatrix}f\\d\\e\\a\\b\\c \end{pmatrix}=\begin{pmatrix}0\\0\\0\\0\\0\\0 \end{pmatrix}$$
It has a nonzero solution, by hypothesis (each line expresses that a certain point belongs to the conic curve).
But a homogeneous system has a nonzero solution iff its determinant is zero.
Remark: Moreover, the equation of this conic is:
$$\det \begin{pmatrix}1 &1 &1 & 1& 1 & \color{red}{1} \\ x_1 & x_2 & x_3 &x_4 & x_5 & \color{red}{x} \\
y_1 & y_2 & y_3 &y_4 & y_5 &\color{red}{y} \\ x_1^2 & x_2^2 & x_3^2 &x_4^2 & x_5^2 & \color{red}{x^2} \\ x_1y_1 & x_2y_2 & x_3y_3 & x_4y_4 & x_5y_5 & \color{red}{xy} \\ y_1^2 & y_2^2 & y_3^2 &y_4^2 & y_5^2 &\color{red}{y^2}
\end{pmatrix}=0$$
