Areas and Special Lines 2

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 $\pm\sqrt{2}$ .

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

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

For the segment $BF$ and the curve $y = (x - 1)^\frac{3}{2} - 1$

$m_{BF} = \dfrac{7\sqrt{2}}{9} \implies y = \dfrac{63\sqrt{2}x - 71\sqrt{2} - 81}{81}$ .

Let $f(x) = \dfrac{63\sqrt{2}x - 71\sqrt{2} - 81}{81}$ and $g(x) = (x - 1)^\frac{3}{2} - 1$

By symmetry the area $A = 2\displaystyle\int_{\frac{11}{9}}^{\frac{17}{9}} f(x) - g(x) \:\ dx = 2\displaystyle\int_{\frac{11}{9}}^{\frac{17}{9}} \dfrac{7}{9}\sqrt{2}x - \dfrac{71\sqrt{2}}{81} - (x - 1)^{\dfrac{3}{2}} dx=$

$2(\dfrac{7\sqrt{2}}{18}x^2 - \dfrac{71\sqrt{2}}{81}x - \dfrac{2}{5}(x - 1)^{\frac{5}{2}})|_{\frac{11}{9}}^{\frac{17}{9}} = \dfrac{44\sqrt{2}}{1215} = \dfrac{\alpha^{\alpha}\beta\sqrt{\alpha}}{\lambda} = \dfrac{2^2 * 11 * \sqrt{2}}{1215} \implies \alpha + \beta + \lambda= \boxed{1228}$