Let A B C be an isosceles triangle in which A is at − 1 , 0 , ∠ A = 2 π 3 , A B = A C and B is on the…

Let, $B\equiv \left(p,0\right)$

$\Rightarrow AB=p+1, p>1$

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

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

$\Rightarrow C\equiv \left(\frac{-p}{2}-\frac{3}{2},\frac{\sqrt{3}}{2}\left(p+1\right)\right)$

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

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

$\Rightarrow p=3$

So, $C\equiv \left(-3,2\sqrt{3}\right)$

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

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

Solving with line $y=x+3$.

$\Rightarrow x+3=\frac{3-x}{\sqrt{3}}$

$\Rightarrow x\sqrt{3}+3\sqrt{3}=3-x$

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

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

$\Rightarrow x=\frac{-3}{2}\left(4-2\sqrt{3}\right)$

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

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

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

$\Rightarrow \frac{{\beta }^4}{{\alpha }^2}=36$