Given two points $A =(x_1,y_1)$, $B = (x_2,y_2)$ and length $L$, how do I plot a parabolic segment of length $L$ that connects A and B? The vertex of the parabola $(p, \, q)$ should be such that $x_1 \leq p \leq x_2$ and $q \leq y_1, y_2$.
In other words, I need to draw a parabolic segment connecting $A$ and $B$ with length $L$ that looks like a 'U' shape, with the vertex being below $A$ and $B$.
Thanks.
On
Suppose we have a parabola defined by $y = a x^2$, with $a \gt 0$. Now we want to know its length between $x_1$ and $x_2$.
$\begin{align} L &= \int dl \\ &= \int \sqrt{dx^2 + dy^2} \end{align}$
$y = a x^2$
$dy = 2 a x dx$
$\begin{align} L &= \int \sqrt{dx^2+ (2 a x dx)^2} \\ &= \int dx \sqrt{1+ (2 a x)^2} \end{align}$
Now we look in a table of integral rules:
$\int \sqrt{1 + (kx)^2} = \frac{1}{k} \sqrt{1 + k^2x^2} + \frac{1}{2} \ln\left(kx + \sqrt{1 + k^2x^2}\right) + C$
Now comes the hard part: given x1, x2 and L, solve for a. This looks like a trancendental equation and offhand I don't see any clever way to solve it analytically. So you'll have to solve it numerically. A simple loop of code should do it; L increases strictly with a.
Shifting to the desired x1, x2, y1 and y2 is trivial (but notice that the problem appears to be underconstrained).
To amplify Beta's answer, if you write the equation as $y=a(x-b)^2+c$ the fact that the two points need to be on the parabola gives two equations in $a,b,c$. Your constraints on the vertex become constraints on $a, b$ and $c$. The arclength from $x_1$ to $x_2$ is $L=\left.\frac{1}{2a}\sqrt{1+4a^2(x-b)^2}+\frac{1}{2}\ln\left(2a(x-b)+\sqrt{1+4a^2(x-b)^2}\right)\right|_{x_1}^{x_2}$ and this is a third equation in $a$ and $b$. It looks like you are into a numeric solution, but you do have enough constraints.