question_answer

If OAB is an equilateral triangle inscribed in the parabola \[{{y}^{2}}=4ax\] with O as the vertex, then the length of the side of the\[\Delta \,\,OAB\] is

A)  \[8\,a\,\sqrt{3}\]         

B)  \[4\,a\,\sqrt{3}\]

C)  \[2\,a\,\sqrt{3}\]          

D)  \[a\,\sqrt{3}\]

Correct Answer: A

Solution :

In \[\Delta OCA,\,\,\,\tan {{30}^{o}}=\frac{AC}{OC}\] \[\Rightarrow \] \[\frac{1}{\sqrt{3}}=\frac{2at}{a{{t}^{2}}}\] \[\Rightarrow \] \[t=2\sqrt{3}\] Again in \[\Delta \,OCA,\] \[OA=\sqrt{O{{C}^{2}}+A{{C}^{2}}}\] \[=\sqrt{{{(a{{t}^{2}})}^{2}}+{{(2at)}^{2}}}\] \[=\sqrt{{{[{{(2\sqrt{3})}^{2}}]}^{2}}{{a}^{2}}+4{{a}^{2}}{{(2\sqrt{3})}^{2}}}\] \[=\sqrt{144\,{{a}^{2}}+48{{a}^{2}}}=\sqrt{192\,{{a}^{2}}}\] \[=8\sqrt{3}\,a\]