TITLE: How to approximate/connect two continuous cubic Bézier curves with/to a single one?
QUESTION [10 upvotes]: I subdivide a cubic Bézier curve at a given t value using de Casteljau’s algorithm, which yields two cubic Bézier curves. Afterwards I “scale” the second curve (proportionally).
I’d like to reconnect or approximate the two curves to/with a single curve in a third step. Is that possible?

This illustrates what I’m intending to do.
I guess reversing de Casteljau’s algorithm won’t work because I don’t have one of the intermediate points.
If there are multiple approaches, I’d favor a simpler (faster to compute) strategy.
Thanks in advance.
Update:
Maybe this figure makes it more clear; it shows all the points I have:

The original cubic Bézier curve is defined by the points $ p_{0}, p_{1}, p_{2}, p_{3} $. 
It is divided at a given $ t $ (timing) value using de Casteljau’s algorithm, which yields the points $ q_{1}, r_{2}, i_{1}, q_{2}, r_{1}, k $ where $ k $ is the division point. 
The two subcurves are defined by the control points $ p_{0}, q_{1}, q_{2}, k $ and $ k, r_{1}, r_{2}, p_{3} $, respectively.
The scaled second subcurve is defined by the points $ k, {r}' _{1}, {r}' _{2}, {r}' _{3} $
Scaling is applied as follows: $ {p}' = k + (p - k) \cdot factor $ for $ r_{1}, r_{2}, p_{3} $

REPLY [2 votes]: You can calculate the distance of q2 k and k r′1  as LeftLength and RightLength.
We can image we have got a cubic curve, which we subdivide at k, but what is the t value?
We can calculate t from de Casteljau’s algorithm, that is t = leftLength / rightLength
We also know that the new curve's p1' is along p0->p1 direction, and p2' is along p3->p2, so we can use t to calculate the new curve's p1' and p2'.