JEE Main 2024 — 1 February, Shift 2 — Question 30

Given $A(-1,0), \angle A = \frac{2\pi}{3} = 120^\circ, AB = AC$ and $B$ lies on the positive x-axis.

Let $AB = AC = s.$

Then,

$$B = (-1 + s, 0)$$

Since AC makes an angle $120^\circ$ with the positive x-axis,

$$\vec{AC} = s(\cos120^\circ, \sin120^\circ) = s\left(-\frac12, \frac{\sqrt3}{2}\right)$$

Hence,

$$C = \left(-1-\frac{s}{2},\; \frac{\sqrt3 s}{2}\right)$$

Given $BC = 4\sqrt3$,

$$BC^2 = \left(\frac{3s}{2}\right)^2 + \left(\frac{\sqrt3 s}{2}\right)^2 = 3s^2$$ $$\Rightarrow \sqrt3\, s = 4\sqrt3 \Rightarrow s = 4$$

Therefore,

$$B=(3,0), \qquad C=(-3,2\sqrt3)$$

Equation of line $BC$:

$$m = \frac{2\sqrt3 - 0}{-3 - 3} = -\frac{\sqrt3}{3}$$ $$y = -\frac{\sqrt3}{3}(x-3)$$

Intersection with the line $y=x+3$:

$$-\frac{\sqrt3}{3}(x-3) = x+3$$ $$\Rightarrow x = 3\sqrt3 - 6$$ $$y = x + 3 = 3\sqrt3 - 3$$

Thus,

$$(\alpha,\beta) = (3\sqrt3 - 6,\; 3\sqrt3 - 3)$$

Required value:

$$\frac{\beta^4}{\alpha^2} = \frac{[3(\sqrt3-1)]^4}{[3(\sqrt3-2)]^2}$$

Using $(\sqrt3-1)^4 = 4(2-\sqrt3)^2$,

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