Finding the vertex of the parabola parameterized by $p(t)=P_0+P_1t+P_2t^2$ for vectors $P_0, P_1, P_2$

Question

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.

Answer 1

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 $$p'(t_\star)\cdot a = 0 \qquad\to\qquad (2at_\star+b)\cdot a = 0 \qquad\to\qquad t_\star = -\frac{a\cdot b}{2a\cdot a}$$

Answer 2

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.