<s>[INST] Parabola passing through TWO points with known tangents directions Given the points $ P_1 (0, 5), P_2 (0, 0), P_3 (3, 0) $, a parabola passes through $P_1$ and is tangent to the segment $P_1 P_2$ and passes through $P_3$ and is tangent to the segment $P_2 P_3$.  Determine the vertex, and axis of symmetry of the parabola, as well the quadratic term coefficient.
My Attempt:
Based on the Bezier quadratic curve, the above parabola can be written as
$P(t) = P_1 (1 - t)^2 + 2 t (1 - t) P_2 + t^2 P_3 $
Expanding this, we get
$P(t) = t^2 (P_1 - 2 P_2 + P_3) + t ( - 2 P_1 + 2 P_2 ) + P_1 $
Note that $P(0) = P_1 $ and $P(1) = P_3$, and that the tangent vector
$P'(t) = (2 t ) (P_1 - 2 P_2 + P_3) + (-2 P_1 + 2 P_2 ) $
so that
$ P'(0) = 2 (P_2 - P_1) $ which is along the segment $P_1 P_2$, and
$ P'(1) = 2 ( P_3 - P_2 ) $ which is along the segment $P_2 P3$
The vertex of of this parabola is the point $P(t_1)$ where $ P'(t_1) \cdot P''(t_1) = 0 $
I still have to find a way to compute the axis of symmetry and the quadratic term coefficient.
EDIT:
The above parabola parametrization can be written as follows
$P(t) = A t^2 + B t + C $
with $A = P_1 - 2 P_2 + P_3 , \ B = -2 P_1 + 2 P_2, \ C = P_1 $
Completing the square in $t$, this becomes
$ P(t) = A' (t - \tau)^2 + B' (t - \tau) + C' $
where $ \tau $ is the value of $t$ where the vertex $C'$ occurs.
Comparing the two expressions which must be equal for all $t$, we have
$ A' = A $
$ - 2 \tau A' + B' = B $
$ A' \tau^2 - B' \tau + C' = C $
which can be solved for $A', B', C'$ if $\tau$ is known. To determine $\tau$, we must use the requirement that $B'$ is perpendicular to $A'$, and this gives
$ A \cdot (B + 2 A \tau) = 0 $
Note that this equation is exaclty the same as $P'(\tau) \cdot P''(\tau) = 0 $
Having found $\tau$ we can readily solve for $A', B', C'$.
Clearly, the axis of symmetry points along vector $A'$ (which is the same as $A$).
If we now choose a unit vector along $B'$ to be our $x'$ axis and a unit vector along vector $A'$, and let the origin $O'$ of this coordinate system be the vector $C'$, then
$P(t) = (x(t), y(t) ) = A' (t - \tau)^2 + B' (t - \tau) + C' $
Define the unit vectors $\hat{u} = \dfrac{B'}{\| B' \|}$ and $\hat{v} = \dfrac{A'}{\| A' \| } $, then
$ \vec{P(t)} = x' \hat{u} + y' \hat{v} + C' $
where $x' = \| B' \| (t - \tau ) $
and $ y' = \| A' \| (t - \tau )^2 $
Hence, $y' = \bigg( \dfrac{\| A' \|}{ \| B' \|^2 } \bigg) x' \ ^2 $
So that the quadratic term coefficient is
$ a =    \dfrac{\| A' \|}{ \| B' \|^2 } $
 [/INST] Following the method outlined in my OP, we have the parametric equation of the parabola,
$P(t) = A t^2 + B t + C = A' (t - \tau)^2 + B' (t - \tau) + C' $
where
$A = P_1 - 2 P_2 + P_3 = (0, 5) - 2 (0, 0) + (3, 0) = (3, 5) $
$B = -2 P_1 + 2 P_2 = - 2 (0, 5) + 2 (0, 0) = (0, -10) $
$C = P_1 = (0, 5) $
And the constant $\tau$ is computed from
$ A \cdot (2 A \tau + B) = 0 $, hence $ (3, 5) \cdot ( (6, 10) \tau + (0, -10) ) = 0 $, which gives $\tau =  \dfrac{50}{68} = \dfrac{25}{34} $
In addition, we have
$A' = A = (3, 5) $
$B' = B + 2 A' \tau = (0, -10) + \dfrac{25}{17} (3, 5) = \dfrac{1}{17} (75, -45) $
$C' = C + B' \tau - A' \tau^2 = (0, 5) + \dfrac{25}{578} (75, -45) - \dfrac{625}{1156} (3, 5) = \dfrac{1}{1156} ( 1875, 405 ) $
$C'$ is the vertex of the parabola.
The unit vectors $\hat{u}$ and $\hat{v}$ are defined as follows
$ \| B' \| = \dfrac{1}{17} \sqrt{7650} = \dfrac{ 15 \sqrt{34} }{17} = \dfrac{30}{\sqrt{34} } $
$ \| A' \| = \sqrt{34} $
$ \hat{u} = \dfrac{B'}{\| B' \|} =  \dfrac{(75, -45) }{\sqrt{7650}} $
$ \hat{v} = \dfrac{A'}{\| A' \|} = \dfrac{ (3, 5) }{\sqrt{34} } $
The quadratic term coefficient is
$ a = \dfrac{ \| A' \|}{\| B' \|^2 } = \dfrac{34 \sqrt{34}}{900}  $
Hence, if $p$ is the semi-latus rectum, then
$ 4 p = \dfrac{1}{a} $
Hence,
$ p = \dfrac{ 225 }{ 34 \sqrt{34} } $
The focus is given by
$F = C' + p \hat{v} = \dfrac{1}{1156} ( 1875, 405 ) + \bigg( \dfrac{ 225 }{ 34 \sqrt{34} } \bigg)  \bigg( \dfrac{ (3, 5) }{\sqrt{34} } \bigg) = \bigg(\dfrac{2550}{1156}, \dfrac{1530}{1156} \bigg) = \bigg( \dfrac{75}{34}, \dfrac{45}{34} \bigg) $
Here's a plot of the parabola showing the vertex, the focus, and the axis of symmetry, as well as the tangency points with the coordinate axes.

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