# Help in finding curve equation.

What I have is length of the bottom line $L$ and area under parabolic curve $S$. How can I find this parabolic curve equation, depending on area under it? The following picture illustrates the problem.

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You should take into account that since you didn't provide the origin, equation is not unique with respect to vertical and horizontal translation. –  Kaster Dec 14 '12 at 9:28
I edited the tags. The question certainly has nothing to do with elliptic curves, and I'm fairly certain that it has nothing to do with integral equations either. Do roll back, if you know otherwise :-) –  Jyrki Lahtonen Dec 14 '12 at 10:03

Assume your equation is $y = ax^2$. Compute the area $S$ through integration to find the area under the curve from $x \in [0,L/2]$. You should be able to find $a$ in terms of $L,S$.

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Let's assume that the parabola is concave and symmetric with respect to $y$ axis. In this case its equation would look like $y(x) = -ax^2+b$, and its graph would be similar to
Obviously, roots are $\{-\sqrt{\frac ba}, \sqrt{\frac ba}\}$. Also $L = 2\sqrt{\frac ba}$, from which $aL^2 = 4b$. Now, let's find the area under the parabola.
$\displaystyle S = 2\int_0^{\frac L2} \left(-ax^2+b\right)dx = \left.\left(-\frac {ax^3}3 + bx\right)\right|_0^{\frac L2} = \frac L2 \left(b-\frac {aL^2}{12}\right)$.
Now we can use the relation $aL^2 = 4b$, so $\displaystyle S = \frac L2 \left(b-\frac b3\right) = \frac {Lb}3$, and finally $\displaystyle b = \frac {3S}L$ and $\displaystyle a = \frac {12S}{L^3}$.
$\displaystyle y(x) = -\frac {12S}{L^3} x^2 + \frac {3S}L$