What's the Special Lines?

Let $y = t^3 - 1 \implies x = t^2 + 1 \implies \dfrac{dy}{dx}|(t = t_{1}) = \dfrac{3}{2}t_{1} \implies$ the tangent line to the curve at $(x(t_{1}),y(t_{1}))$ is: $y - (t_{1}^3 - 1) = \dfrac{3}{2}t_{1}(x - (t_{1}^2+ 1))$

Let the line be normal to the curve at $(x(t_{2}),y(t_{2})) \implies (t_{2} - t_{1})(t_{2}^2 + t_{1}t_{2} + t_{1}^2) = \dfrac{3}{2}t_{1}(t_{2} - t_{1})(t_{2} + t_{1}) \implies \dfrac{1}{2}(t_{2} - t_{1})(2t_{2}^2 - t_{1}t_{2} - t_{1}^2) = 0$ $t_{1} \neq t_{2} \implies t_{2} = -\dfrac{t_{1}}{2}$

Since the tangent is also normal to the curve at $(x(t_{2}),y(t_{2})) \implies \dfrac{9}{4}t_{1}t_{2} = -1$ $\implies \dfrac{9}{8}t_{1}^2 = 1 \implies t_{1} = \pm\dfrac{2\sqrt{2}}{3} \implies$ the two slopes are $\pm\sqrt{2}$ .

slope $= \sqrt{2}$ and $t_{1} = \dfrac{2\sqrt{2}}{3} \implies y_{0} - \sqrt{2}x_{0} = \dfrac{-35\sqrt{2} - 27}{27}$

slope $= -\sqrt{2}$ and $t_{2} = \dfrac{\sqrt{2}}{3} \implies y_{0} + \sqrt{2}x_{0} = \dfrac{35\sqrt{2} - 27}{27}$

$\implies y_{0} = -1$ and $x_{0} = \dfrac{35}{27} \implies x_{0} + y_{0} = \dfrac{35}{27} - 1 = \dfrac{8}{27} = \dfrac{a}{b} \implies a + b = \boxed{35}$ .