Calculate midpoint of a parabola having the start and end tangent I have the start and end points of a line segment, and a start tangent and an end tangent. This represents a second degree edge (of a finite element mesh) modelled by a parabola. That means that there is one extra point in between. This point is the projection of the middle point of the edge in a parabola that passes through these two points with the given start and end tangents.
I need the position of this middle point. This is a problem in 3d space. Is there only one point that satisfies these constraints? Any help?
Here's a picture:
Image
Thank you.
 A: Given a parabola in a Euclidean plane, the interior is a convex figure. Thus, any line is (a) contained in the exterior with no intersection, or (b) is tangent to and intersects the parabola once at the point of tangency, or (c) parallel to the main axis and intersects in only one point, or (d) intersects the parabola in only two points.
We are given two points $A,B$ which are points of tangency of a parabola and their tangent lines. They must intersect in a point $C$. Construct $D$ as the midpoint of line segment $AC$, $E$ as the midpoint of line segment $BC$, $F$ as the midpoint of line segment $DE$ ,and $G$ as the midpoint of line segment $AB$. Now $F$ is a point of tangency of the parabola. This construction of a new tangent point and line of a parabola can be continued with tangent point pairs $A,F$ or $B,F$ and so on iteratively. Note that $F$ is also the midpoint of line segment $CG$.
The three points $A,F,B$ determine the parabola uniquely. We can now find the unique intersection of the parabolic arc between the points $A,B$ with the perpendicular bisector of line segment $AB$ which intersects $AB$ at $G$. Of course, this can all be expressed algebraically using vectors.
A: You have enough information to construct a quadratic Bézier patch for this piece of the parabola: Let $P_2$ be the intersection of the two tangent lines. Then the arc of the parabola between $P_0$ and $P_1$ can be parameterized as $$\begin{align} P(\lambda) &= (1-\lambda)[(1-\lambda)P_0+\lambda P_2]+\lambda[(1-\lambda)P_2+\lambda P_1] \\ &= \lambda^2(P_0-2P_2+P_1)+2\lambda(P_2-P_0)+P_0\end{align}\tag1$$ with $0\le\lambda\le1$. The point you’re looking for is the intersection of this parabolic arc with the plane perpendicular to the line segment $\overline{P_0P_1}$ through its midpoint $M=\frac12(P_0+P_1)$. This plane has the equation $$(P_1-P_0)\cdot P=(P_1-P_0)\cdot M.\tag2$$ Combining equations (1) and (2) yields a quadratic equation in $\lambda$, which I’m sure you know how to solve.  
In practice, computing $P_2$ can be problematic. Small perturbations to either of the tangent lines can cause them not to intersect. There are various ways to deal with this, for example, by computing the intersections of their projections onto the coordinate planes and then mapping those back to the lines or doing some sort of averaging.
