I need some help with this.
Given $3$ points (and assuming the $y$-axis is the vertical axis):
$A (x1, y1, z1) - B (x2, y2, z2)$ and vertex $(x3, y3, z3) where:
$$x3 = \frac{x_1+x_2}{2}$$
$$y3 = \frac{y_1+y_2}{2} + \frac{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}{2}$$
$$z3 = \frac{z_1+z_2}{2}$$
Can I, given this data, get a standard parabola through these $3$ points in a vertical plane?
What I need is the following:
Three functions, $x(t), y(t)$ and $z(t)$, where $t(0)$ would result in point $A$, and $t(s)$ would result in point $B$.
Basically, the vertex is always directly in between $A$ and $B$, where the height is the average of $A$ and $B$ plus half of the distance between them (So basically one can draw a perfect circle through these $3$ points and the center of the circle would be perfectly in between $A$ and $B$ (shown in blue on the image)), see the image below for an example projected in $2D$ space.
I think the formula you wanted for $y_3$ is
$$y_3 = \frac{y_1+y_2}{2} + \frac{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}}{2}.$$
If a "standard parabola" means a parabola whose axis is parallel to one of the coordinate axes, and if the vertex of the parabola must be at the point labeled "Vertex" in the diagram, then the answer is no, you cannot construct such a parabola except in the case where $y_1 = y_2.$
If the vertex of the parabola is at the point labeled "Vertex" and the axis of the parabola is parallel to the $y$ axis, then the axis of the parabola is the vertical line in the diagram. Two points on the parabola at the same "height" ($y$ value) will be at the same distance from the axis; two points at different heights will be at different distances. (The point at the greater height will be closer to the axis.) Since the points $A$ and $B$ are at equal distances from the axis, the only way they can both be on the parabola is if they are also at the same height, that is, if $y_1 = y_2.$
If you merely require that the parabola have a vertical axis and pass through the three points, but do not require $(x_3,y_3,z_3)$ to be the vertex of the parabola, you can construct the parabola, but the vertex will be outside the circle.
If $x_1 \neq x_2,$ one way to do this is to consider only the $x$ and $y$ coordinates initially. Let $$ y_0 = y_3 - \frac{y_1+y_2}{2}. $$ A parabola that passes through the $(x,y)$ points $(x_1,0),$ $(x_2,y_0),$ and $(x_3,y_3)$ is given by the formula $$ y = y_0 - \frac{y_0}{(x_1 - x_3)^2}(x - x_3)^2.$$
But we want the parabola to pass through points $A$ and $B,$ which are on the line given by $$ y = y_1 + \frac{y_2 - y_1}{x_2 - x_1} (x - x_1). $$
So we add the two formulas on the right-hand side of these equations: $$ y = y_0 + y_1 + \frac{y_2 - y_1}{x_2 - x_1} (x - x_1) - \frac{y_0}{(x_1 - x_3)^2}(x - x_3)^2. \tag1$$
If desired, you can expand and recombine the terms to put this in the general form $y = ax^2 + bx + x.$
But the parabola needs to be in the vertical plane that goes through $A$ and $B.$ That means that $z$ is a function of $x$. In particular,
$$ z = z_1 + \frac{z_2 - z_1}{x_2 - x_1} (x - x_1). \tag2 $$
Equations $(1)$ and $(2)$ together completely describe the parabola.
Note that we started with a parabola with a vertex at $(x_3,y_0,z_3)$, which is on the vertical line in the diagram, but when we added $y_1 + \frac{y_2 - y_1}{x_2 - x_1} (x - x_1)$ to the formula, if $y_2\neq y_1$ we shift the vertex away from that vertical line and outside the circle.
Still, if the line $AB$ is not too steep, the vertex will not be far outside the circle. Even if $AB$ is almost vertical, there is a limit to how high the vertex can be above the circle, and that limit is only a fraction of the circle's radius.
If you do not require the axis of the parabola to be vertical, there is a parabola through all three points with the given vertex.