JEE MAIN - Mathematics (2024 - 1st February Evening Shift - No. 26)
Let $A B C$ be an isosceles triangle in which $A$ is at $(-1,0), \angle A=\frac{2 \pi}{3}, A B=A C$ and $B$ is on the positve $x$-axis. If $\mathrm{BC}=4 \sqrt{3}$ and the line $\mathrm{BC}$ intersects the line $y=x+3$ at $(\alpha, \beta)$, then $\frac{\beta^4}{\alpha^2}$ is __________.
Answer
36
Explanation
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$\frac{\mathrm{c}}{\sin 30^{\circ}}=\frac{4 \sqrt{3}}{\sin 120^{\circ}}[$ By sine rule $]$
$$ 2 c=8 \Rightarrow c=4 $$
$\begin{gathered}\mathrm{AB}=|(\mathrm{b}+1)|=4 \\\\ \mathrm{~b}=3, \mathrm{~m}_{\mathrm{AB}}=0 \\\\ \mathrm{~m}_{\mathrm{BC}}=\frac{-1}{\sqrt{3}} \\\\ \mathrm{BC}:-\mathrm{y}=\frac{-1}{\sqrt{3}}(\mathrm{x}-3) \\\\ \sqrt{3} \mathrm{y}+\mathrm{x}=3\end{gathered}$
Point of intersection : $y=x+3, \sqrt{3} y+x=3$
$\begin{aligned} & ({\sqrt{3}+1}) y=6 \\\\ & y=\frac{6}{\sqrt{3}+1} \\\\ & x=\frac{6}{\sqrt{3}+1}-3 \\\\ & =\frac{6-3 \sqrt{3}-3}{\sqrt{3}+1} \\\\ & =3 \frac{(1-\sqrt{3})}{(1+\sqrt{3})}=\frac{-6}{(1+\sqrt{3})^2}\end{aligned}$
$\frac{\beta^4}{\alpha^2}=36$
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