Given a cubic bezier curve,
$P(t) = P_0(1-t)^3 + 3P_1t(1-t)^2 + 3P_2 t^2(1-t) + P_3 t^3$
where $P(t) = (x(t), y(t))$ and $P_i = (x_i, y_i)$.
$P_0$ and $P_3$ are end points and $P_1$ and $P_2$ are control points.
Assume that $P_0 = (0,0)$ and $P_3 = (K,K)$ for some $K \in \mathbb{R}^+$. $P_1$ and $P_2$ are constrained to lie in the square defined by $P_0$ and $P_3$ as opposite ends of its diagonal.
I have the $x(t)$ coordinate, say $x^*$. The problem is to find an explicit equation for the corresponding $y(t)$ in terms of $x^*$. This means that I want a function $g$ in $x$ only such that $y^* = g(x^*)$.
I am hoping that with the constraints on the control points an explicit equation can be derived for $y$ given $x^*$.
EDIT: We know $P_1$ and $P_2$.
Thanks.
Best Answer
I think deletoin $t$ beween $$x=3x_1t(1-t)^2+3x_2t^2(1-t)+kt^3$$ and $$y=3y_1t(1-t)^2+3y_2t^2(1-t)+kt^3$$ is very difficult, for power 3. As you see here
with changing $y_1$ and $y_2$ we have a family of curves which give us a differential equation. I think finding an equation of $x(t)$, $y(t)$, $y_1$ and $y_2$ with fixed $x_1$ and $x_2$ is impossible. In this case by deletion $y_1$ or $y_2$, differential equation is of first order and if delete both of them, so differential equation is of second order. Fnding Envelope maybe useful here.