Can a cubic Bezier curve be a quadratic one if two control points are equal from the cubic one? I have been struggling to find out a solution for a case where i have a cubic Bezier curve where two arbitrary control points of the one are equal, therefore i should show that this curve can be the quadratic curve. Do you have any thoughts how to prove it?
 A: The cubic Bezier curve with control points $A$, $B$, $C$, $D$ has equation
$$
P(t) = (1-t)^3A + 3t(1-t)^2B + 3t^2(1-t)C + t^3D
$$
We can re-arrange this to get
$$
P(t) = (-A + 3B -3C +D)t^3 + (3A-6B+3C)t^2 + (-3A+3B)t + A
$$
So, the cubic curve will become quadratic if and only if $-A + 3B -3C +D = 0$.
This condition can also be written as
$$
D - A = 3(C-B)
$$
This has a nice geometric meaning: it says that the vector $\overrightarrow{AD}$ is three times the vector $\overrightarrow{BC}$.
As you can see, if two control points are equal, this does not mean that the curve will be quadratic.
A: Scroll down in http://caffeineowl.com/graphics/2d/vectorial/bezierintro.html to the title "Transforming a quadratic Bezier in a cubic Bezier". There it is mentioned how the handles $C_1$ and $C_2$ of a cubic curve are to be positioned, when the handle $C$ of a quadratic Bezier curve together with the points $P_1$, $P_2$ are given and the cubic one should output the same result. 
In fact you have
$C_1 = 2/3•C + 1/3•P_1$ and $C_2 = 2/3•C + 1/3•P_2$.
(And this is essentially different from your assumption $C=C_1=C_2$!)
--- rk
