Let ABC be an isosceles triangle in which A is at (- 1, 0), ∠ A =...

Let, $B ≡ \left(p, 0\right)$

$⇒ AB = p + 1, p > 1$

Now, finding point $C$ using parametric form we get,

$\dfrac{x + 1}{cos\dfrac{2\pi}{3}} = \dfrac{y - 0}{sin\dfrac{2\pi}{3}} = p + 1$

$⇒ C ≡ \left(\dfrac{- p}{2} - \dfrac{3}{2}, \dfrac{\sqrt{3}}{2}\left(p + 1\right)\right)$

$⇒ B{C}^{2} = {\left(\dfrac{- 3}{2}\left(p + 1\right)\right)}^{2} + {\left(\dfrac{\sqrt{3}}{2}\left(p + 1\right)\right)}^{2}$

$⇒ BC = \sqrt{3}\left(p + 1\right) = 4\sqrt{3}$

$⇒ p = 3$

So, $C ≡ \left(- 3, 2\sqrt{3}\right)$

Now, equation of line $BC$ is given by,

$y = \dfrac{- 1}{\sqrt{3}}\left(x - 3\right)$

Solving with line $y = x + 3$.

$⇒ x + 3 = \dfrac{3 - x}{\sqrt{3}}$

$⇒ x\sqrt{3} + 3\sqrt{3} = 3 - x$

$⇒ x\left(\sqrt{3} + 1\right) = 3\left(1 - \sqrt{3}\right)$

$⇒ x = \dfrac{3\left(1 - \sqrt{3}\right)}{1 + \sqrt{3}}$

$⇒ x = \dfrac{- 3}{2}\left(4 - 2\sqrt{3}\right)$

$⇒ \alpha = - 3\left(2 - \sqrt{3}\right)\text{and}\beta = - 3 + 3\sqrt{3}$

$⇒ \dfrac{{\beta}^{4}}{{\alpha}^{2}} = \dfrac{81{\left(\sqrt{3} - 1\right)}^{4}}{9{\left(2 - \sqrt{3}\right)}^{2}}$

$⇒ \dfrac{{\beta}^{4}}{{\alpha}^{2}} = \dfrac{9{\left(4 - 2\sqrt{3}\right)}^{2}}{{\left(2 - \sqrt{3}\right)}^{2}}$

$⇒ \dfrac{{\beta}^{4}}{{\alpha}^{2}} = 36$