QJEEM24Ma7183: JEE Main 2024 Mathematics - Class 11 JEEM24Ma7183 Solution: Let be an isosceles triangle in which is at and is on the…

Question

Verified Solution • 2025–26 Session moderate

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 __________.

Accepted Answer

\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} &amp; ({\sqrt{3}+1}) y=6 \\\\ &amp; y=\frac{6}{\sqrt{3}+1} \\\\ &amp; x=\frac{6}{\sqrt{3}+1}-3 \\\\ &amp; =\frac{6-3 \sqrt{3}-3}{\sqrt{3}+1} \\\\ &amp; =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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