Parabola passing through TWO points with known tangents directions

Question

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 } $

Answer 1

This can also be solved in a purely geometric fashion.

Let $M=(3/2, 5/2)$ be the midpoint of $P_1P_2$ and $V=(3/4, 5/4)$ the midpoint of $P_2V$. Then point $V$ is on the parabola and line $P_2M$ is parallel to the axis.

In this case finding focus and directrix is very simple, because perpendicular tangents meet on the directrix and their contact chord passes through the focus. Hence the directrix is the line through $P_2$ perpendicular to $P_2M$ and the focus $F$ is the second intersection between $P_1P_3$ and the circle of centre $V$ and radius $VP_2$: $$ F=\left({75 \over 34}, {45 \over 34}\right). $$ From that one can find the axis, vertex $A$ and quadratic coefficient: $\displaystyle{1\over4AF}={34\sqrt{34}\over900}$.

Answer 2

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.