Are the two endpoints you have at the same height so the arch is symmetric? That is, are the two endpoints (a, c) and (b, c)? You can simplify a little by letting d= (a-b)/2, half of the distance from a to b and then looking at a parabola from (-d, 0) to (d, 0). Then you know that the and the fact that y= 0 when x= d means so that .(all these pararmeters it is me that decide them), what i think is if you know the starting and end points and you the length, knowing that is can only assume a specific shape you only have two solutions, one is the mirror of the other, however i cant define or the radius or the arch equation, i am hoping someone can enlighten me about this.
Now the hard part. If , then and the "differential of arclength" is . The arclength, from 0 to d, is . To do that, let so that . Also . When x= 0, . When x= d, . The arclength is .
That, I think, can be done by a trig substitution followed by "partial fractions". In any case, set the arclength (from -d to d so twice that integral) equal to whatever the length is to be and solve for a and b.