The area (in sq. units) of an equilateral tri | vedclass

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(C) Let the parabola be $y^2 = 8x$. The vertex is at $O(0,0)$.Let the equilateral triangle be $OAB$, where $A$ and $B$ lie on the parabola.Let the coo

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$192$

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(C) Let the parabola be $y^2 = 8x$. The vertex is at $O(0,0)$.Let the equilateral triangle be $OAB$, where $A$ and $B$ lie on the parabola.Let the coordinates of $A$ be $(2t^2, 4t)$ for some $t > 0$.Since the triangle is equilateral and symmetric about the $x$-axis, the coordinates of $B$ are $(2t^2, -4t)$.The side length of the triangle is $AB = 4t - (-4t) = 8t$.The angle $\angle AOx = 30^{\circ}$ because the triangle is equilateral and the $x$-axis bisects the angle at the vertex.Thus, $	an 30^{\circ} = \frac{4t}{2t^2} = \frac{2}{t}$.Since $	an 30^{\circ} = \frac{1}{\sqrt{3}}$, we have $\frac{1}{\sqrt{3}} = \frac{2}{t}$, which gives $t = 2\sqrt{3}$.The side length $s = 8t = 8(2\sqrt{3}) = 16\sqrt{3}$.The area of an equilateral triangle is given by $\frac{\sqrt{3}}{4} s^2$.Area $= \frac{\sqrt{3}}{4} (16\sqrt{3})^2 = \frac{\sqrt{3}}{4} (256 	imes 3) = \frac{\sqrt{3}}{4} (768) = 192\sqrt{3}$ sq. units.

The area (in sq. units) of an equilateral triangle inscribed in the parabola $y^{2}=8x$, with one of its vertices at the vertex of this parabola, is (in $\sqrt{3}$)

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