Is there a way to determine if all points of a bezier curve are visible from an endpoint? For instance, if you're given a cubic bezier curve in the plane: $\textbf C(t) = \sum\_{i=0}^3 B\_i^3(t) \textbf P^i$ would any of the rays from $\textbf P^0$ to $\textbf C(t)$ intersect the curve at any other point? 


So I want to know if the following has 0 or >=1 #solutions:


all the rays from the origin: $\textbf L(s) := s { \textbf a}$


$\textbf C(0) = \textbf L(0) = {\textbf 0}$


$\textbf C(t1) = \textbf L(s1) = s1 { \textbf a}$


$\textbf C(t2) = \textbf L(s2) = s2 { \textbf a}$


this appears to have 4 equations in 5 unknowns (${ \textbf a}$ can be normalized)?


where $B\_i^3$ is the Bernstein basis: $B\_i^3(t) = C(3,i)(1-t)^{3-i}t^i$ 


thanks!



None or infinite #solns for Bezier Curve Problem?
WLOG take ${\bf P}^0 = (0,0)$ and ${\bf P}^3 = (1,0)$, and let ${\bf P}^1 = (x\_1, y\_1)$ and ${\bf P}^2 = (x\_2, y\_2)$. For $t \ne 0$ the tangent to the curve at ${\bf C}(t)$ is parallel to ${\bf C}(t)-{\bf P}^0$ iff $$ \left( -3x\_{{2}}y\_{{1}}+3x\_{{1}}y\_{{2}}-y\_{{2}}+2y\_{{1}}
 \right) {t}^{2}-\left( -6x\_{{2}}y\_{{1}}+2y\_{{1}}+6x\_{{1}}y\_{{2
}} \right) t-3x\_{{2}}y\_{{1}}+3x\_{{1}}y\_{{2}} = 0$$
The points will all be visible if this quadratic (as a function of $t$) has no roots in $(0,1)$ or a double root there; if there is at least one simple root in the interval not all points will be visible.


