Hard Algebra Problem

$\begin{cases} x = y^4\\ x + y^2 = 1 \end{cases}$

Substituting,

$y^4 + y^2 - 1= 0\\ y^2 = \dfrac{-1 \pm \sqrt 5}2\text{(reject negative root)}\\ y^2 = \dfrac{\sqrt 5 - 1}2\\ y = \pm\sqrt{\dfrac{\sqrt 5 - 1}2}$

Since $x + y^2 = 1$,

$x = 1 - \dfrac{\sqrt 5 - 1}2 = \dfrac{3 -\sqrt 5}2$

So the intersection points are $\left(\dfrac{3 - \sqrt 5}2, \pm\sqrt{\dfrac{\sqrt 5 - 1}2}\right)$.

The distance is $2 \sqrt {\dfrac{\sqrt 5 - 1}2} = \sqrt{2(\sqrt 5 - 1)} = \sqrt{-2 + 2 \sqrt 5}$.

Then $(u,v) = (-2, 2)$.