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An equilateral triangle is inscribed in the parabola $y^{2}=4 x$ one of whose vertex is at the vertex of the parabola, the length of each side of the triangle is
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The correct answer is:
$8 \sqrt{3}$
Let $A B=\ell$, then $A M=\ell \cos 30^{\circ}=\frac{\ell \sqrt{3}}{2}$ $\& \mathrm{BM}=\ell \sin 30^{\circ}=\frac{\ell}{2}$
So, the coordinates of $\mathrm{B}$ are $\left(\frac{\ell \sqrt{3}}{2}, \frac{\ell}{2}\right)$ Since, $\mathrm{B}$ lies on $\mathrm{y}^{2}=4 \mathrm{x}$
$\begin{array}{l}
\therefore \frac{\ell^{2}}{4}=4\left(\frac{\ell \sqrt{3}}{2}\right) \\
\Rightarrow \ell^{2}=\frac{16}{2} \cdot \sqrt{3} \ell \Rightarrow \ell=8 \sqrt{3}
\end{array}$
So, the coordinates of $\mathrm{B}$ are $\left(\frac{\ell \sqrt{3}}{2}, \frac{\ell}{2}\right)$ Since, $\mathrm{B}$ lies on $\mathrm{y}^{2}=4 \mathrm{x}$
$\begin{array}{l}
\therefore \frac{\ell^{2}}{4}=4\left(\frac{\ell \sqrt{3}}{2}\right) \\
\Rightarrow \ell^{2}=\frac{16}{2} \cdot \sqrt{3} \ell \Rightarrow \ell=8 \sqrt{3}
\end{array}$
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