Bezier Curve Problem, finding missing control point Given the two sets of control points: A: $(1, 2)$, $(2, 3)$, $(a, b)$, $(4, 2)$.
B: $(4, 2)$, $(c, d)$, $(5, 5)$, $(6, 4)$.
Find values for the control points $(a, b)$ and $(c, d)$ so that the resulting composite Bezier curve is smooth at the control point $(4, 2)$.
How do I even go about starting this? 
A: For the composite curve to be smooth at $(4\mid 2)\,$, the three control points $(a\mid b)\,$, $(4\mid 2)\,$, $(c\mid d)\,$ must be collinear --
i.e., $\begin{vmatrix}
a&b&1\\
4&2&1\\
c&d&1
\end{vmatrix}=0\,$.
A: @Senex Ægypti Parvi It depends of which smoothness we are speaking. 
If it is a parametrization smoothness, i.e. not only the direction of the tangent but also the intensity of the speed vector should be identical, we must have a stronger constraint (2 conditions instead of 1):
$(4,2)$ should be the midpoint of $(a,b)$ and $(c,d)$, i.e., $\dfrac{a+c}{2}=4$ and  $\dfrac{b+d}{2}=2$.
A: Denoting the two set of control points as $A_i$ and $B_i$ ($i=0$~$3$), we can compute the 2nd derivatives at $t=1$ of curve A and at $t=0$ for curve B as 
$A^"(1)=6(A_1-2A_2+A_3)$
$B^"(0)=6(B_0-2B_1+B_2)$
We also assume that the continuity at point $(4,2)$ is $C^2$ as $C^1$ continuity cannot result in an unique solution for $(a,b)$ and $(c,d)$. Then, we have $A^"(1)=B^"(0)$, from which we get
$\begin{cases}6-2a=4-2c \\ 5-2b=7-2d \end{cases}$ 
We also have $C^1$ continuity at point $(4,2)$, which will lead to 
$\begin{cases}a+c=8\\ b+d=4 \end{cases}$ 
So, in the end we can solve for $(a,b)=(3.25, 1.5)$ and $(c,d)=(4.75,2.5)$. 
