Let ABC be an isosceles triangle in which A is at (-1,0),angle A= 2π 3,AB=AC and B is on the positive x-axis. If BC=4√3 and the line BC intersects…

Let, $B\equiv (p,0)$B≡(p,0)

$$\Rightarrow AB=p+1,p>1$$⇒AB=p+1,p>1

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

$$\frac{x+1}{\mathrm{cos}\frac{2\pi }{3}}=\frac{y-0}{\mathrm{sin}\frac{2\pi }{3}}=p+1$$cos32π​x+1​=sin32π​y−0​=p+1

$$\Rightarrow C\equiv (\frac{-p}{2}-\frac{3}{2},\frac{\sqrt{3}}{2}(p+1))$$⇒C≡(2−p​−23​,23​​(p+1))

$$\Rightarrow B{C}^{2}={(\frac{-3}{2}(p+1))}^{2}+{(\frac{\sqrt{3}}{2}(p+1))}^{2}$$⇒BC2=(2−3​(p+1))2+(23​​(p+1))2

$$\Rightarrow BC=\sqrt{3}(p+1)=4\sqrt{3}$$⇒BC=3​(p+1)=43​

$$\Rightarrow p=3$$⇒p=3

So, $C\equiv (-3,2\sqrt{3})$C≡(−3,23​)

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

$$y=\frac{-1}{\sqrt{3}}(x-3)$$y=3​−1​(x−3)

Solving with line $y=x+3$y=x+3.

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

$$\Rightarrow x\sqrt{3}+3\sqrt{3}=3-x$$⇒x3​+33​=3−x

$$\Rightarrow x(\sqrt{3}+1)=3(1-\sqrt{3})$$⇒x(3​+1)=3(1−3​)

$$\Rightarrow x=\frac{3(1-\sqrt{3})}{1+\sqrt{3}}$$⇒x=1+3​3(1−3​)​

$$\Rightarrow x=\frac{-3}{2}(4-2\sqrt{3})$$⇒x=2−3​(4−23​)

$$\Rightarrow \alpha =-3(2-\sqrt{3})\text{and}\beta =-3+3\sqrt{3}$$⇒α=−3(2−3​)andβ=−3+33​

$$\Rightarrow \frac{{\beta }^{4}}{{\alpha }^{2}}=\frac{81{(\sqrt{3}-1)}^{4}}{9{(2-\sqrt{3})}^{2}}$$⇒α2β4​=9(2−3​)281(3​−1)4​

$$\Rightarrow \frac{{\beta }^{4}}{{\alpha }^{2}}=\frac{9{(4-2\sqrt{3})}^{2}}{{(2-\sqrt{3})}^{2}}$$⇒α2β4​=(2−3​)29(4−23​)2​

$$\Rightarrow \frac{{\beta }^{4}}{{\alpha }^{2}}=36$$⇒α2β4​=36