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(B) Let the vertices of the equilateral triangle be $O(0, 0)$,$P(x, y)$,and $Q(x, -y)$.Since the triangle is equilateral and symmetric about the $x$-a

Vedclass · Accepted Answer

$8a\sqrt{3}$

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(B) Let the vertices of the equilateral triangle be $O(0, 0)$,$P(x, y)$,and $Q(x, -y)$.Since the triangle is equilateral and symmetric about the $x$-axis,the angle made by the side $OP$ with the $x$-axis is $30^\circ$.The equation of the line $OP$ is $y = x 	an(30^\circ) = \frac{x}{\sqrt{3}}$,or $x = y\sqrt{3}$.Substituting this into the parabola equation $y^2 = 4ax$,we get $y^2 = 4a(y\sqrt{3}) = 4\sqrt{3}ay$.Since $y 
eq 0$,we have $y = 4\sqrt{3}a$.Then $x = (4\sqrt{3}a)\sqrt{3} = 12a$.The side length $L$ of the triangle is the distance $PQ$,which is $2y = 2(4\sqrt{3}a) = 8a\sqrt{3}$.

An equilateral triangle is inscribed in the parabola $y^2 = 4ax$ such that one of its vertices is at the origin $(0, 0)$ and the other two vertices lie on the parabola. The length of its side is equal to

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