Equation of a parabola in 3D space I have two points with coordinates A(x1,y1,z1) and B(x2,y2,z2). There is a third point which is vertex(lowest point) of the parabola. I only know z-coordinate of this point. I need to find coordinates of the points that lie on a parabola that passes through these 3 points. I am interested in coordinates that lie in between A and B.
 A: According to your comment added later the parabola lies in a vertical plane through the two points $A=(x_0,y_0,z_0)$ and $B=(x_1,y_1,z_1)$ and has a vertical axis. It therefore has a parametric representation of the form
$$s\mapsto\left\{\eqalign{x(s)&=(1-s)x_0+sx_1 \cr y(s)&=(1-s)y_0+sy_1 \cr z(s)&=as^2+bs+c\cr}\right.$$
It remains to determine the three coefficients $a$, $b$, $c$ from the data. We want $z(0)=z_0$ and $z(1)=z_1$. The third equation comes from the fact that we are also given the minimal value $z_*$ of $s\mapsto z(s)=as^2+bs+c$. This leads to the equation
$$-{b^2\over 4a}+c=z_*\ ,$$
so that we can now determine $a$, $b$, $c$ in a straightforward way.
A: Here is a "dinosaur" method for three points: (1,2,3) , (-1,4,2) and (2,1,3) and let the parametrics look like $x=at^2+bt+c, y=dt^2+et+f, z=gt^2+ht+i$ Substitution of the three points respectively gives: $at^2+bt+c=1, dt^2 +et +f = 2, gt^2+ht+i=3$ and $at^2+bt+c=-1, dt^2 +et +f = 4, gt^2+ht+i=2$ and $at^2+bt+c=2, dt^2 +et +f = 1, gt^2+ht+i=3$ In the first set I let $t=1$, in the second set $t=-1$ and in the third set $t=2$ to obtain three system of three equations. The first one would be $a+b+c=1, a-b+c=-1, 4a+2b+c=2$. You verify the other two. Solving these systems gives the following: $a=0 , b=1 , c=0 , d=0 , e=-1 , f=3 , g=-1/6 , h=0.5 , i=8/3$ Now use a 3D plot program on your computer to feed in these parametric equations and verify that it is of parabolic nature, indeed passing through my three given points.
