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  • Area of an equilateral triangle inscribed in the parabola   with one of its vertices on the vertex of this parabola is:  O

Area of an equilateral triangle inscribed in the parabola y^2 = 8x  with one of its vertices on the vertex of this parabola is: 
Option: 1 128\sqrt3
Option: 2 192\sqrt3
Option: 3 64\sqrt3
Option: 4 256\sqrt3

Answers (1)

best_answer

Let the two vertices of the triangle be Q and R. Points Q and R will have the same x - coordinate =k( say )

Now in the right triangle PRT, right-angled at T.

\tan 30^{\circ}=\frac{\mathrm{RT}}{\mathrm{k}} \Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{RT}}{\mathrm{k}} \Rightarrow \mathrm{RT}=\frac{\mathrm{k}}{\sqrt{3}} \Rightarrow \mathrm{R}\left(\mathrm{k}, \frac{\mathrm{k}}{\sqrt{3}}\right)

Now R lies on the parabola: y^{2}=4 ax

\\\Rightarrow\left(\frac{k}{\sqrt{3}}\right)^{2}=4 a(k) \\ \\\Rightarrow \frac{k}{3}=4 a \\ \\\Rightarrow k=12 a

\\\text { Length of side of the triangle }=2(\mathrm{RT})=2 \cdot \frac{\mathrm{k}}{\sqrt{3}}=2 \cdot \frac{(12 \mathrm{a})}{\sqrt{3}}=8 \sqrt{3} \mathrm{a}=16\sqrt{3}

\text { Area of Equilateral triangle is }= \frac{\sqrt{3}}4a^2 =192\sqrt{3}

 

Posted by

Suraj Bhandari

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