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An equilateral triangle is inscribed in the parabola   \mathrm {y^2=4 a x}  whose vertex is at the vertex of the parabola. Find the length of its side.
 

Option: 1

\mathrm {4 a \sqrt{3}}


Option: 2

\mathrm {8 a \sqrt{2}}


Option: 3

\mathrm {8 a \sqrt{3}}


Option: 4

\mathrm {16 a \sqrt{3}}


Answers (1)

best_answer

We know that each angle of an equilateral triangle is 

60^{\circ} . A$ is $(0,0)$ and $\mathrm {A B=l},  then


\begin{aligned} & \mathrm {x=A M=l \cos 30^{\circ}=l \cdot \sqrt{3} / 2 .} \\ & \mathrm {y=B M=l \sin 30^{\circ}=l / 2 .} \end{aligned}

\mathrm {Also \: \: y^2=4 a x }

\mathrm {\frac{l^2}{4}=4 a \cdot \frac{\sqrt{3}}{2} \cdot l \quad \therefore l=8 a \sqrt{3} . }

 

Posted by

chirag

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