$\newcommand{\C}{\mathbb{C}}$ $\newcommand{\im}{\operatorname{Im}}$ Let the rational normal cone be given by the image of $\Phi: \C^2 \to \C^{d+1}$, parametrized by $(s,t) \to (s^d, s^{d-1}t,...t^d)$.
One can show that this is given by ideal generated by the $2 \times 2$-minors of $\psi_x=\begin{pmatrix} x_0 & x_1&...& x_{d-1} \\ x_1 & x_2 & ... & x_{d} \end{pmatrix} $.
Thus this is a determinantal variety.
My question is 'What is the linear map $\begin{pmatrix} x_0 & x_1&...& x_{d-1} \\ x_1 & x_2 & ... & x_{d} \end{pmatrix} $?'. That is where would it have come from?
It seems likely that the map $\begin{pmatrix} x_0 & x_1&...& x_{d-1} \\ x_1 & x_2 & ... & x_{d} \end{pmatrix}$ comes from somewhere: The reason is with any determinantal variety, the ideal of the variety is always given by the condition of the kernel of some linear map being of maximal rank. This condition here is (by working backwards we know that the answer has to be this) that the maximal rank of the kernel of the map $\phi_x=1$ gives the condition of the minors vanishing.