Dancing On a Parabola

The answer is $-\sqrt{2} = \boxed{-1.414214}$ .

The gradient of the parabola at point $A$ is: $\quad m_P = \dfrac {dy}{dx} = 2x = 2a$

Therefore, the gradient of line segment $L$ which is perpendicular to the parabola at point $A$ is: $\quad m_L = -\dfrac {1}{m_P} = -\dfrac {1}{2a}$

$\Rightarrow \dfrac {a^2-b^2}{a-b} = a+b = -\dfrac {1}{2a}\quad \Rightarrow b = -a - \dfrac {1}{2a}$

$\begin{aligned} \Rightarrow L^2 & = (a^2-b^2)^2+(a-b)^2 = \left( -\dfrac {a-b}{2a} \right)^2 +(a-b)^2 \\ & = (a-b)^2 \left( \dfrac {1}{4a^2} + 1 \right) = \left( 2a+\dfrac{1}{2a} \right)^2 \left( \dfrac {1}{4a^2} + 1 \right) \\ & = \dfrac {(4a^2+1)^3}{16a^4} \end{aligned}$

To find the minimum $L$ ,

$\dfrac {dL^2}{da} = \dfrac {3(4a^2+1)^2(8a^2) - 4(4a^2+1)^3}{16a^5}$

When $L$ is minimum, $24a^2-16a^2-4=0\quad \Rightarrow a^2= \dfrac{1}{2} \quad \Rightarrow a = \dfrac{1}{\sqrt{2}}$

When $b = -\dfrac{1}{\sqrt{2}} - \dfrac{\sqrt{2}}{2} = - \dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}} = \boxed{-\sqrt{2}}$