# 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 Option: 2 Option: 3 Option: 4

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: $y^{2}=4 ax$

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

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

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

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