Axis of Symmetry of a Parabola

A line connecting midpoint of tangency points $M(3,-1.5)$ and the intersection point of tangency lines $I(-0.75, -1.25)$ is parallel to the axis of symmetry. The slope of this line is $\dfrac{-1}{15}$ . The angle $A=arctan(-1/15)\approx -3.814$ .

We can get easily the parabola equation via the Bèzier curve equation

$\displaystyle c(t)=\sum_{k=0}^n {2\choose k}t^k(1-t)^{2-k} \vec{p}_k$ , $t\in\mathbb{R}$

where $\vec{p}_0 = (5,-7)$ , $\vec{p}_1 = (\displaystyle-\frac{3}{4},-\frac{5}{4})$ and $\vec{p}_2 = (1,4)$ and $n=2$ . Point $\vec{p}_1$ is the result of intersection between the two lines. By solving the equation we get $c(t)$

$\displaystyle c(t) = (\frac{15}{2}t^2-\frac{23}{2}t+5,-\frac{1}{2}t^2+\frac{23}{2}t-7)$ .

Now, $c(t)$ can be written as

$\displaystyle c(t)=\begin{bmatrix} 5 \\ -7 \\ \end{bmatrix} + \begin{bmatrix} -\frac{23}{2} & \frac{15}{2} \\ \frac{23}{2} & -\frac{1}{2}\\ \end{bmatrix} \begin{bmatrix} t \\ t^2 \\ \end{bmatrix}$

namely as transformation of the $(t,t^2)$ parabola in the $xy$ plane. Due to the fact that the axis of $(t,t^2)$ is $(0,t)$ , the axis $a(t)$ of $c(t)$ is given by the application

$\displaystyle a(t)=\begin{bmatrix} 5 \\ -7 \\ \end{bmatrix} + \begin{bmatrix} -\frac{23}{2} & \frac{15}{2} \\ \frac{23}{2} & -\frac{1}{2}\\ \end{bmatrix} \begin{bmatrix} 0 \\ t \\ \end{bmatrix} = (\frac{15}{2}t+5,-\frac{1}{2}t-7)$

or, explicitly

$\displaystyle y=-\frac{x}{15}-\frac{100}{15}$

Eventually

$\displaystyle\left|\arctan\left(-\frac{1}{15}\right)\right|=3.814°$

A conic has a general equation of $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ , and a parabola has the property $B^2 - 4AC = 0$ . Since $(1, 4)$ and $(5, -7)$ are solutions to the conic, we have $A + 4B + 16C + D + 4E + F = 0$ and $25A - 35B + 49C + 5D - 7E + F = 0$ .

The derivative of the conic is $2Ax + B(x\frac{dy}{dx} + y) + 2Cy\frac{dy}{dx} + D + E\frac{dy}{dx} = 0$ , and since $\frac{dy}{dx} = 3$ at $(1, 4)$ and since $\frac{dy}{dx} = -1$ at $(5, -7)$ we have $2A + 7B + 24C + D + 3E = 0$ and $10A - 12B + 14C + D - E = 0$ .

Solving these $5$ equations gives $(A, B, C, D, E, F) = (A, \frac{8}{11}A, \frac{16}{121}A, -\frac{54}{11}A, -\frac{216}{121}A, \frac{729}{121}A)$ and $(A, B, C, D, E, F) = (A, 30A, 225A, -3503A, -703A, 2594A)$ .

The first solution degenerates to a line in the form of $y = -\frac{11}{4}x + \frac{27}{4}$ , but the second solution gives us a parabola in the form of $x^2 + 30xy + 225y^2 - 3503x - 703y + 2594 = 0$ , and its angle of rotation is $\theta = \frac{1}{2}\arctan(\frac{B}{A - C}) =\frac{1}{2}\arctan(\frac{30}{1 - 225}) \approx \boxed{-3.814}$ .

Parabola parametrization:

$\large{\vec{P} = \vec{P_0} + \alpha \, \vec{v_1} + \beta \, \alpha^2 \, \vec{v_2} \\ x = x_0 + \alpha \, v_{1 x} + \beta \, \alpha^2 \, v_{2 x} \\ y = y_0 + \alpha \, v_{1 y} + \beta \, \alpha^2 \, v_{2 y} }$

In the above equation, $P_0$ is the vertex, $\alpha$ is a variable spatial parameter, $\beta$ is a scaling parameter, and vectors $v_1$ and $v_2$ are orthogonal unit vectors.

$\large{\vec{v_1} = (cos \theta, sin \theta) \\ \vec{v_2} = (-sin \theta, cos \theta)}$

There are two points, and thus two $\alpha$ values to find. The six unknowns are thus:

$\large{x_0 \\ y_0 \\ \theta \\ \beta \\ \alpha_1 \\ \alpha_2}$

There are six equations (four for position and two for slope). First, some definitions:

$\large{(x_1,y_1) = (1,4) \\ (x_2,y_2) = (5,-7) \\ s_1 = 3 \\ s_2 = -1}$

Some elaboration on the slope equation derivations:

$\large{\frac{dy}{dx} = \frac{dy/d \alpha}{dx / d \alpha} = \frac{v_{1 y} + 2 \beta \, \alpha \, v_{2 y}}{v_{1 x} + 2 \beta \, \alpha \, v_{2 x}}}$

Form a system of six non-linear equations to solve for the six unknowns:

$\large{x_0 + \alpha_1 \, cos \theta - \beta \, \alpha_1^2 \, sin\theta - x_1 = 0 \\ y_0 + \alpha_1 \, sin \theta + \beta \, \alpha_1^2 \, cos\theta - y_1 = 0 \\ x_0 + \alpha_2 \, cos \theta - \beta \, \alpha_2^2 \, sin\theta - x_2 = 0 \\ y_0 + \alpha_2 \, sin \theta + \beta \, \alpha_2^2 \, cos\theta - y_2 = 0 \\ s_1 (cos \theta - 2 \beta \, \alpha_1 \, sin \theta) - (sin \theta + 2 \beta \, \alpha_1 \, cos \theta) = 0 \\ s_2 (cos \theta - 2 \beta \, \alpha_2 \, sin \theta) - (sin \theta + 2 \beta \, \alpha_2 \, cos \theta) = 0 }$

Solving with multivariate Newton Raphson yields (note that the axis angle leads $\theta$ by 90 degrees):

x0
0.60858328765

y0
2.031404182

theta (degrees)
-93.8140748343

beta
0.0655361479401

alpha1
-1.99027234142

alpha2
8.7192883529

Axis angle (degrees)
-3.81407483429