Areas and 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 $x_{0} = \dfrac{35}{27}$ and the two tangents intersect at $A:(\dfrac{35}{27},-1)$ .

$B:(\dfrac{17}{9},\dfrac{16\sqrt{2} - 27}{27})$

$C:(\dfrac{17}{9},,\dfrac{-16\sqrt{2} - 27}{27})$

$D: (\dfrac{17}{9},-1)$

$AD = \dfrac{16}{27}$ and $BC = \dfrac{32\sqrt{2}}{27}$

$\implies$ the area $A_{2}$ of $\triangle{ABC}$ is $A_{2} = \dfrac{1}{2}(BC)(AD) = \dfrac{256}{729}\sqrt{2}$ .

$F: (\dfrac{11}{9}, \dfrac{2\sqrt{2} - 27}{27})$

$E: (\dfrac{11}{9}, \dfrac{-2\sqrt{2} - 27}{27})$

The midpoint of $EF$ is $G:(\dfrac{11}{9},-1)$

$GD = \dfrac{2}{3}$ and $FE = \dfrac{4\sqrt{2}}{27}$

$\implies$ the area $A_{1}$ of trapezoid $CEFB$ is $A_{1} = \dfrac{1}{2}GD(FE + BC) = \dfrac{4\sqrt{2}}{9}$

$\implies A = A_{1} - A_{2} = \dfrac{68}{729}\sqrt{2} = \dfrac{\alpha}{\beta}\sqrt{2} \implies \alpha + \beta = \boxed{797}$ .