jmc

To find the $y$-coordinates of the intersections, substitute $y^4$ for $x$ in $x+y^2=1$ and solve for $y$, resulting in $$\begin{array}{rl}{y}^4+y^2& =1\\ \Rightarrow y^4+y^2-1& =0\\ \Rightarrow y^2& =\frac{-1\pm \sqrt{1+4}}{2}=\frac{-1\pm \sqrt{5}}{2}\end{array}$$But $y^2$ is positive, so we reject $\frac{-1-\sqrt{5}}{2}$. Therefore $y=\pm \sqrt{\frac{\sqrt{5}-1}{2}}$.

Using each of these coordinates to solve for $x$ gives us the intersections at $\left(\frac{3-\sqrt{5}}{2},\sqrt{\frac{\sqrt{5}-1}{2}}\right)$ and $\left(\frac{3-\sqrt{5}}{2},-\sqrt{\frac{\sqrt{5}-1}{2}}\right)$. Using the distance formula, we have

$$\begin{array}{rl}& \sqrt{{\left(\frac{3-\sqrt{5}}{2}-\frac{3-\sqrt{5}}{2}\right)}^2+{\left(\sqrt{\frac{\sqrt{5}-1}{2}}+\sqrt{\frac{\sqrt{5}-1}{2}}\right)}^2}\\ & =\sqrt{{\left(2\sqrt{\frac{\sqrt{5}-1}{2}}\right)}^2}\\ & =2\sqrt{\frac{\sqrt{5}-1}{2}}\\ & =\sqrt{2\sqrt{5}-2}.\end{array}$$So, $(u,v)=\boxed{(-2,2)}.$