Finding points on $\vec r(t) = \vec At^3 + \vec Bt^2 + \vec Ct + \vec D$ where the tangent is parallel to the line $px + qy + k = 0$

Question

Given a line defined by the equation: $$px + qy + k = 0$$ and a parametric cubic curve defined by: $$\vec r(t) = \vec At^3 + \vec Bt^2 + \vec Ct + \vec D$$ where both curves lie in 2D space, how can I find all points on the cubic curve $t$ where the line is parallel to the tangent in these points?

Answer 1

Considering that $\vec r(t) = (x(t),y(t))$ and that the line

$$ p x + q y + k = 0 $$

can be written as $\vec r(t) = p_0 + \vec v t$ or $\cases{x = x_0 + v_x t\\ y = y_0 + v_y t}$ with the relationship between $p_0,\vec v, p, q, k$ which follows (assuming $v_x\ne 0$ and $v_y\ne 0$)

$$ \frac{x-x_0}{v_x}=\frac{y-y_0}{v_y}\Rightarrow v_y x- v_x y - x_0y_y+y_0 v_x = 0 $$

then $p = v_y, q = -v_x, k = - x_0y_y+y_0 v_x$

so we follow with $\vec r_1(t) = \vec At^3 + \vec Bt^2 + \vec Ct + \vec D$ and $\vec r_2(t) = p_0 + \vec v t$

Now the tangent vectors to $\vec r_1(t)$ are given by $\vec r_1'(t) = 3\vec At^2 + 2\vec Bt + \vec C$ and the tangent vectors to $r_2(t)$ are given by $\vec r_2'(t) = \vec v$. Finally if $\vec r_1'(t)$ and $\vec r_2'(t)$ are parallel then

$$ \vec r_1'(t) \times \vec r_2'(t) = \vec 0 $$

etc.