The search
I needed to convert some quadratic interpolation plots (from PiCTeX) to quadratic Bézier splines (for SVG). I saw this question asked several times elsewhere but never with the kind of answer I was looking for. I finally just did my own calculations, and in the process also discovered how to do handle the cubic case.
The quadratic case
The interplolation plots are splines (piecewise
interpolations), each segment of which is specified by three points: two
endpoints and a midpoint. The endpoints of adjoining segments coincide.
PiCTeX computes and plots a quadratic curve through these three points using
Lagrangian interpolation.
Quadratic Bézier splines are also in segments specified by three points: two endpoints and a middle control point.
The difference between these two is in the middle point. In the interpolation spline, the midpoint is on the curve, but in the Bézier spline, the middle point is off the curve (unless the "curve" is a straight line).
Call the endpoints $a$ and $b$, the interpolation midpoint $m$, and the Bézier middle control point $q$. Using the parameterization $0 \le T \le 1$, the Lagrangian quadratic interpolation $P(T)$ through points $a$, $m$, $b$ has $P(0) = a$, $P(\frac12) = m$, $P(1) = b$, and
$$\eqalign{ P(T) &= \frac{(T-\frac12)(T-1)}{(-\frac12)(-1)} a + \frac{T(T-1)}{(\frac12)(-\frac12)} m + \frac{T(T-\frac12)}{1(\frac12)} b \cr &= (2T^2 - 3T + 1) a - 4(T^2 - T) m + (2T^2 - T) b }$$
The corresponding quadratic Bézier curve, where $q$ is the middle control point, is
$$\eqalign{ B(T) &= (1-T)^2 a + 2(1-T)T q + T^2 b \cr &= (1-2T+T^2) a + 2(T-T^2) q + T^2 b }$$
At $T = \frac12$, these are
$$\eqalign{ P({\textstyle\frac12}) &= m \cr B({\textstyle\frac12}) &= \frac14 a + \frac12 q + \frac14 b }$$
If $P(\frac12) = B(\frac12)$ then
$$ m = \frac14 a + \frac12 q + \frac14 b $$
Key result: Solving the previous equation for $q$ gives
$$ q = - \frac12 a + 2 m - \frac12 b $$
Possibly unnecessary verification: The above shows that these two curves have three points in common. Since they are both quadratic functions of $T$, it may be unnessary to further verify that the two curves coincide. But I did this anyway by substituting this value for $q$ into $B(T)$ and finding it always equals $P(T)$:
$$\eqalign{ B(T) &= (1-2T-T^2) a + 2(T-T^2) \left(-\frac12 a + 2 m - \frac12 b \right) + T^2 b \cr &= (2T^2-3T+1) a - 4(T^2-T) m + (2T^2-T) b \cr &= P(T) }$$
The cubic case
Cubic interpolation plots have two midpoints $m$ and $n$, and cubic Bézier plots have two middle control points $p$ and $q$. Using conventions analogous to those above, we have
$$\eqalign{ P(T) = &\ \frac{(T-\frac13)(T-\frac23)(T-1) }{(-\frac13)(-\frac23)(-1 )} a + \frac{ T (T-\frac23)(T-1) }{( \frac13)(-\frac13)(-\frac23)} m \cr &+ \frac{ T (T-\frac13)(T-1) }{( \frac23)( \frac13)(-\frac13)} n + \frac{ T (T-\frac13)(T-\frac23)}{ 1 ( \frac23)( \frac13)} b \cr = &- \frac92 \left( T^3 - 2 T^2 + \frac{11}9 T - \frac29 \right) a + \frac{27}2 \left( T^3 - \frac53 T^2 + \frac23 T \right) m \cr &- \frac{27}2 \left( T^3 - \frac43 T^2 + \frac13 T \right) n + \frac92 \left( T^3 - T^2 + \frac29 T \right) b \cr B(T) =&\ (1-T)^3 a + 3(1-T)^2T p + 3(1-T)T^2 q + T^3 b \cr =&\ (-T^3+3T^2-3T+1) a + 3(T^3-2T^2+T) p + 3(-T^3+T^2) q + T^3 b }$$
At $T = \frac13$ and $\frac23$, these are
$$\eqalign{ P({\textstyle\frac13}) &= m \cr B({\textstyle\frac13}) &= \frac8{27} a + \frac49 p + \frac29 q + \frac1{27} b \cr P({\textstyle\frac23}) &= n \cr B({\textstyle\frac23}) &= \frac1{27} a + \frac29 p + \frac49 q + \frac8{27} b }$$
If $P(\frac13) = B(\frac13)$ and $P(\frac23) = B(\frac23)$ then
$$\eqalign{ m &= \frac8{27} a + \frac49 p + \frac29 q + \frac1{27} b \cr n &= \frac1{27} a + \frac29 p + \frac49 q + \frac8{27} b }$$
Then
$$\eqalign{ -2m + n &= - \frac{15}{27} a - \frac69 p + \frac{ 6}{27} b \cr m - 2n &= \frac{ 6}{27} a - \frac69 q - \frac{15}{27} b }$$
Key result: Solving the previous system for $p$ and $q$ gives
$$\eqalign{ p &= - \frac56 a + 3 m - \frac32 n + \frac13 b \cr q &= \frac13 a - \frac32 m + 3 n - \frac56 b }$$
Possibly unnecessary verification: The above shows that the two cubic curves have four points in common, which may be enough to show that they coincide, but subsituting the values for $p$ and $q$ into $B(T)$ explicitly shows that it always equals $P(T)$:
$$\eqalign{ B(T) = &\ (-T^3+3T^2-3T+1) a \cr &+ 3 (T^3-2T^2+T) \left( -\frac56a + 3m - \frac32n + \frac13b \right) \cr &+ 3 (-T^3+T^2) \left( \frac13a - \frac32m + 3n - \frac56b \right) \cr &+ T^3 b \cr = &\ \left( -\frac92 T^3 + 9 T^2 - \frac{11}2 T + 1 \right) a \cr &+ \left( \frac{27}2 T^3 + \frac{45}2 T^2 + 9 T \right) m \cr &+ \left( -\frac{27}2 T^3 + \frac{36}2 T^2 - \frac92 T \right) n \cr &+ \left( \frac92 T^3 - \frac92 T^2 + T \right) b \cr = &P(T) }$$
This doesn't answer the question. In fact, as far as I can see, there isn't a question. But, anyway ...
The equation $$ m = \frac14 a + \frac12 q + \frac14 b $$ can be written as $$ m = \frac12 q + \frac12\left( \frac12 a + \frac12 b \right) = \frac12 q + \frac12 c $$ where $c = \tfrac12(a+b)$ is the mid-point of the triangle side $ab$. So, $m$ is the mid-point of a median of the triangle $aqb$. Along the same lines, we have $$ q = 2m - c = c+ 2(m-c) $$ which has a similar geometric interpretation.
You can do the same sorts of things with cubics, but the geometric pictures are not quite as nice.
The cubic case is covered in detail in this book by Mortenson.