Parabolic arclength Considering a parabola of the form $y=ax^2+bx$ I was curious what relation $a$ and $b$ would need to have in order for the arclength of the parabola between its zeroes to be twice the distance between its zeroes. 
To simplify the calculus, I set the integral from the first zero to the vertex equal to the distance between the zeroes. 
$-\frac{b}{a}=\int_0^{-\frac{b}{2a}}\sqrt{1+(2ax+b)^2}dx$  
Which leads to (I believe)
$4b = \left[ (2ax+b)\sqrt{(2ax+b)^2+1} + \ln{(2ax+b+\sqrt{(2ax+b)^2+1})} \right]_0^{-\frac{b}{2a}}$
Thing is, with those limits of integration, you get 
$4b = b\sqrt{b^2+1} + \ln{(b + \sqrt{b^2+1})}$
Which not only have I not been able to solve for $b$, it doesn't have an $a$ in it, and since the total arc length between the zeroes must be at least the distance between them, I.e. $L>-\frac{b}{2a}$, I find it hard to believe $a$ is not in the formula. 
Sorry for the incomplete work, I'm on my phone, I can upload a picture of the complete work soon. 
 A: This looks to be correct.
Considering a parabola of the form $y=ax^2+bx+c$, we have $$\int\sqrt{1+(2ax+b)^2}\,dx=-\frac{b\sqrt{b^2+1} +\sinh ^{-1}(b)}{4 a}$$ provided $\Re(b)\neq 0\lor -1\leq \Im(b)\leq 1$ and then $$\int_\alpha^\beta\sqrt{1+(2ax+b)^2}\,dx=\frac{\alpha\sqrt{\alpha ^2+1}  +\sinh ^{-1}(\alpha )-\beta  \sqrt{\beta ^2+1}-\sinh
   ^{-1}(\beta )}{4 a}$$ Because of the specific bounds you gave, parameter $a$ disappears and, as you obtained, $$L=\int_0^{-\frac{b}{2a}}\sqrt{1+(2ax+b)^2}\,dx=-\frac{b\sqrt{b^2+1}+\sinh ^{-1}(b)}{4 a}$$ So, the equation is effectively $$F(b)=b \left(\sqrt{b^2+1}-4\right)+\sinh ^{-1}(b)$$ and, beside the trivial $b=0$ already excluded by the calculation of the antiderivative, it has two roots which cannot be made explicit; there numerical values are $\pm 3.269202312$.
Inverse symbolic calculators did not find anything corresponding to this number except an estimate $$b_*=\pm(1+e^{\frac{\sqrt{1+e^{\phi }}}{3}})\approx \pm 3.269202317$$ which seems to be quite close (for this value, $f(b_*) \approx 1.2149\times 10^{-8}$) and I suppose that we could keep this nice expression as a good solution. In fact, if we use Newton method at this point, the change for the first iterate is $\Delta b\approx 4.2817\times 10^{-9}$.
