Given a Bezier curve $\gamma$(t) defined by 3 control points P0 = (-1,4), P1 = (0, 0), P2 = (1, 0) such that the curve lies on the parabola $\ y = (x-1)^2 $. Reparameterize to $\alpha$(t) = $\gamma$(3t-1) in standard basis.
Firstly how would I create $\gamma$(t) with respect to the control points. Ignoring the control points I get $\gamma(t) = (t, (t-1)^2) = (0,1) + (1, -2)t + (0,1)t^2$ which respects the parabola $\ y = (x-1)^2 $. But this $\gamma(t)$ is not respective to the control points??
Secondly to reparametrize do I just trivially replace t with 3t-1?
ATTEMPT 2:
$\gamma(t) = (2t-1,(2t-2)^2) $ this is respective of the control points where $\gamma(0) = P0$ and $\gamma(1) = P2$
So does the reparameterization to $\alpha(t) = (-3,16) + (6,-48)t + (0,36)t^2 $ ?
You want to find $A$, $B$, $C$ such that $Ax^2 + Bx + C = y$, so solve:
$$\begin{bmatrix} (-1)^2 & -1 & 1 \\ 0^2 & 0 & 1 \\ 1^2 & 1 & 1 \end{bmatrix} \begin{bmatrix} A \\ B \\ C \end{bmatrix} = \begin{bmatrix} 4 \\ 0 \\ 0 \end{bmatrix}$$
If the statement $\alpha(t) = \gamma(3t - 1)$ is to be taken literally, then yes.