Finding the vertex of the parabola parameterized by $p(t)=P_0+P_1t+P_2t^2$ for vectors $P_0, P_1, P_2$ A parabola can always be described in parametric form by position vector $p(t)$,
$p(t) = P_0 + P_1 t + P_2 t^2 $
where $P_0, P_1, P_2$ are vectors in $2D$ or $3D$.
I would like to prove that the vertex of this parabola is given by
$ V = P_0 + P_1 t_1 + P_2 t_1^2 $
where $t_1$ is the solution of
$p'(t_1) \cdot p''(t_1) = 0 $
which means that $t_1$ is the solution of
$ (P_1 + 2 P_2 t) \cdot (P_2) = 0 $
therefore,
$ t_1 = -\dfrac{ P_1 \cdot P_2 }{ 2 P_2 \cdot P_2 } $
My attempt:
Let's express $t$ as $t = s + t_1$, where $s$ is the new parameter and $t_1$ is the fixed value we want to derive.  Substituting this
$p = P_0 + P_1 (s + t_1) + P_2 (s+t_1)^2\\
  = P_0 + P_1 t_1 + P_2 t_1^2 + (P_1 + 2 P_2 t_1 ) s + P_2 s^2 $
Now, if we select $t_1$ such that $(P_1 + 2 P_2 t_1) \perp P_2$ then
$p = V + u_1 s + u_2 s^2 $
where $u_1 \perp u_2 $ , which clearly implies that $V$ is the vertex, because then the normalized $u_1, u_2$ would constitute a rotation matrix $R = [\hat{u_1}, \hat{u_2}]$, and we would have
$ p = V + R D [s , s^2 ]^T $
where $D = \text{diag}( \|u_1\|, \|u_2\| ) $, so that if $(x', y')$ is the coordinate of $(p - V)$ in the basis $R$ then
$ x' = \| u_1 \| s $ and $y' = \|u_2 \| s^2 $
thus $(x',y') $ is on the parabola $y' = a x'^2 $ where $a = \dfrac{ \| u_2 \| }{\| u_1 \|^2 }$
And this parabola has its vertex at $(x',y') = (0,0)$ at which we have
$p = V = P_0 + P_1 t_1 + P_2 t_1^2 $
Now, to make $(P_1 + 2 P_2 t_1) \perp P_2$, we set
$(P_1 + 2 P_2 t_1) \cdot P_2 = 0$
and this results in $t_1 = - \dfrac{P_1 \cdot P_2}{2 P_2 \cdot P_2} $
Any additional help on this proof or alternative proofs is appreciated.
 A: According to proposition VI of Newton's Principia, the radius of curvature is given by
\begin{equation}
R = \frac{\|V\|^3}{\|V\times A\|}
\end{equation}
where $V = \frac{d p}{d t}$ is the velocity, $A = \frac{d V}{d t}$ is the acceleration.
In our case, $V = P_1 + 2 P_2 t$ and $A = 2 P_2$ which means that $\|V\times A\|$ is a constant equal to $2 \|P_1 \times P_2\|$. We assume that $P_1$ and $P_2$ are independent vectors.
It follows that the radius of curvature is minimal when $\|V\|$ is minimal, which implies $V\cdot A = 0$, hence
\begin{equation}
(P_1 + 2 P_2 t_1)\cdot 2 P_2 = 0
\end{equation}
This gives the vertex of the parabola.
A: Let the vectors be $a$, $b$, $c$ so that the parabola is parameterized as
$$p(t) = a t^2 + b t + c$$
Note that chords with endpoints $p(t)$ and $p(-t)$ are parallel to $b$, and their midpoints form a line in the direction of $a$:
$$p(t)-p(-t)=2bt \qquad\qquad \frac12\left(\,p(t)+p(-t)\,\right)= at^2+ c$$

We "know" that the parabola's axis is parallel to the line of midpoints of any family of parallel chords, hence it is parallel to $a$. Finally, since the tangent at the vertex is perpendicular to the axis (hence, perpendicular to $a$), we need only solve
$$p'(t_\star)\cdot a = 0 \qquad\to\qquad (2at_\star+b)\cdot a = 0 \qquad\to\qquad t_\star = -\frac{a\cdot b}{2a\cdot a}$$
