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

• A

  $\frac{\sqrt{3}}{2}$
• B

  $4 \frac{\sqrt{3}}{2}$
• C

  $8 \frac{\sqrt{3}}{2}$
• Correct answer
  D

  $8 \sqrt{3}$

Solution

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 $\mathrm{y}^{2}=4 \mathrm{x}$ $\begin{array}{l} \therefore \frac{\ell^{2}}{4}=4\left(\frac{\ell \sqrt{3}}{2}\right) \\ \Rightarrow \ell^{2}=\frac{16}{2} \cdot \sqrt{3} \ell \Rightarrow \ell=8 \sqrt{3} \end{array}$

Hence, the correct answer is option: 'D'.