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An equilateral triangle is inscribed in the parabola \mathrm{y^{2} = 4ax}, such that one vertex of this triangle coincides with the vertex of the parabola. Side length of this triangle is -

Option: 1

\mathrm{4a\sqrt3}


Option: 2

\mathrm{6a\sqrt3}


Option: 3

\mathrm{2a\sqrt3}


Option: 4

\mathrm{8a\sqrt3}


Answers (1)

best_answer

If triangle \mathrm{OAB} is equilateral then \mathrm{OA}=\mathrm{OB}=\mathrm{AB}. Thus \mathrm{AB} will be a double ordinate of the parabola.
Thus \mathrm{\angle \mathrm{AOX}=\angle \mathrm{XOB}=\frac{\pi}{6}.}

Let  \mathrm{\mathrm{A}=\left(\mathrm{at}_1{ }^2, 2 \mathrm{at}\right)\: then \: \mathrm{B} \equiv\left(\mathrm{at}_1{ }^2, 2 \mathrm{at}_1\right)}

\mathrm{ \mathrm{m}_{\mathrm{OA}}=\frac{2}{\mathrm{t}_1}=\frac{1}{\sqrt{3}} }

\mathrm{ \Rightarrow \quad \mathrm{t}_1=2 \sqrt{3} \Rightarrow \mathrm{AB}=4 \mathrm{at}_1=8 \mathrm{a} \sqrt{3}\: units. }

Hence option 4 is correct.

 

Posted by

Riya

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