If one of the vertices of an equilateral triangle inscribed in the parabola $y^2=12 x$ coincides with the vertex of the parabola, then the area (in sq. units) of that triangle is

• A

  $192 \sqrt{3}$
• B

  $864 \sqrt{3}$
• C

  $216 \sqrt{3}$
• Correct answer
  D

  $432 \sqrt{3}$

Solution

$y^2=12 x$ $\begin{aligned} & \Rightarrow \quad a^2 \sin ^2 30^{\circ}=12 \cdot a \cos 30^{\circ} \\ & \Rightarrow \quad a=48 \cos 30^{\circ}=24 \sqrt{3} \\ & \text { Area }=\frac{\sqrt{3}}{4} a^2=\frac{\sqrt{3}}{4}(24 \sqrt{3})^2=432 \sqrt{3} .\end{aligned}$

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