Q. 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 $\mathrm{\Delta}OAB$ is
A) $8a\sqrt{3}$
B) $4a\sqrt{3}$
C) $2a\sqrt{3}$
D) $a\sqrt{3}$
Explanation
(a)
In $\mathrm{\Delta}OCA,\ \ \tan{30{^\circ} = \frac{AC}{OC}}$
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$\Rightarrow \ \ \frac{1}{\sqrt{3}} = \frac{2at}{at^{2}},\ t = 2\sqrt{3}$
Again in $\mathrm{\Delta}OCA,$
$OA = \ \sqrt{OC^{2} + AC^{2}} = \sqrt{\left( at^{2} \right)^{2} + (2at)^{2}}$
$= \sqrt{\left\lbrack \left( 2\sqrt{3} \right)^{2} \right\rbrack^{2}a^{2} + 4a^{2}\left( 2\sqrt{3} \right)^{2}} = \sqrt{192a^{2}} = 8a\sqrt{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.