Finding the function of a parabolic curve between two tangents Alright folks, first question here so let me make the situation and background clear.
I'm attempting to start studying for aerospace engineering, so I'm working on improving on my math skills as we speak, but couldn't help "jumping in" to some of the design and starting to look at it. I'm not adverse to algebra, although my calculus legs haven't been walked on in awhile, which is why I haven't been able to get farther myself on my own question.
Another side reason is that I'm attempting to get the gist of equation-based curves in CAD, so rocket engine nozzle curves are perfect for learning that...if I can figure out the equations!
If the format or my thoughts seem a bit off, I have an idea of the concepts in play here, my math skills have just atrophied a little too much for my own comfort.
What I'm working with is the G.V.R. Rao approximation of a bell nozzle curve; essentially, 
$$
f_c = \left\{ 
  \begin{array}{l l}
    0.382R_t & \quad \text{For divergent throat curve}\\
    f_p & \quad \text{Main body}
  \end{array} \right.
$$
Where $f_p$ starts from a point with a tangent of angle $\theta_n$ and ends at a point with a tangent of angle $\theta_e$. $f_p$ also has to fit in a region equal to $L_f - 0.382R_t$, where $L_f$ is the complete distance between the throat and exit plane, so the displacement in the x- or y-axis, depending on how you view the rocket (orientation-wise).
I do know how to differentiate the curve $f_p$ to get $f'_p$ and then find the angle of the slope at a point, but this is backsolving from two slopes to find the region in between.
If it's any help, ideally I'd be constructing the nozzle in CAD vertically, that is, $y_n > y_e$.
What I'm looking for is help toward the derivation of a formula that allows me to construct a curve that is smooth between the two points. One of the reasons I've had a hard time figuring out the exact parameters is because it feels a lot like curve-fitting, which I haven't had much experience with.
If anyone can help break it down for me, it'd be much appreciated but if the question turns out to be too vague, references to places where I can get the requisite learning would be also appreciated.
 A: I think that I understand that you're doing this in 2D. So I'll tell you how to think about the problem of finding a parametric curve $g(t) = (at^2 + bt + c, dt^2 + et + f)$ that satisfies four conditions:
\begin{align}
g(0) &= A \\
g(1) &= B \\
g'(0) &= v \\
g'(1) &= w
\end{align}
The short answer: for arbitrary $A, B, v, w$, this has no solution. If $A$ and $B$ have the same $x$-coordinate, and $v$ and $w$ have the same $y$ coordinate, for instance, it's generally impossible to solve. Why is this? Think about finding $a, b, c$: those are three unknowns, but you're placing 4 conditions on them. In general, there will be no solution. Let's see the details. Writing out everything in just the $x$-coordinates, we get
\begin{align}
c &= A_x \\
a+b+c &= B_x \\
b &= v_x \\
2a+b &= w_x
\end{align}
which I got by evaluating the $x$-coordinate of $g$ at $t = 0$, i.e., $a 0^2 + b \cdot 0 + c$, and setting it to the $x$-coordinate of $A$, and similarly for the other 4. Doing a little shuffling, we get 
\begin{align}
c &= A_x \\
b &= v_x \\
a+b+c &= B_x \\
2a+b &= w_x
\end{align}
and a substitution in the second equation gives
\begin{align}
c &= A_x \\
b &= v_x \\
a &= B_x - A_x - v_x \\
2a+b &= w_x
\end{align}
The final equation says that $2a + b$ must be $w_x$. But the first three tell us that $2a + b$ is 
$$
2a + b = 2(B_x - A_x - v_x) + v_x = 2B_x - 2A_x - v_x
$$
So if we're given a problem in which
$$
2B_x - 2A_x - v_x \ne w_x
$$
then there's no hope of solving it. 
The good news? If you're willing to use cubics, you CAN solve it, and the solution is unique. The technique is pretty much the same as what I just did here. Pretty much any computer graphics book will describe this under "Bezier curves", although I have a fondness for "Computer Graphics: Principles and Practice, 3rd ed", of which I'm the lead author. 
A: Suggesting a numerical approach. Start with a special conic in plane $$ a x^2+ 2  h xy + b y^2 + c x +d y +1 =0,h^2=ab, $$ as for a tilted parabola ( eccentricity $e$=0) in a simpler radical form
$$ y(x) = ax+b +\sqrt {cx+d}$$
and eliminate four constants by repeated differentiation:
$$ y' =a + \frac{c}{2\sqrt {cx+d}}$$
$$ y'' =  \frac{-c^2/4}{{(cx+d)}^{3/2}}$$
$$ y''' =  \frac{(3/8)c^3 }{{(cx+d)}^{5/2}}$$
$$ y''''=  \frac{(-15/16) c^4/2 }{{(cx+d)}^{7/2}}$$
leading to differential equation
$$ 5 \,y^{''} y^{''''}= 3 \,y^{'''2}$$
which can be solved using four $y,y^{'}$ values at start and at the last points as a boundary value problem. A problem that I had solved ca 1970.
