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(A) Let $OAB$ be the equilateral triangle inscribed in the parabola $y^{2}=4ax$,where $O$ is the origin $(0,0)$.Let $AB$ be a chord perpendicular to t

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$8\sqrt{3}a$

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(A) Let $OAB$ be the equilateral triangle inscribed in the parabola $y^{2}=4ax$,where $O$ is the origin $(0,0)$.Let $AB$ be a chord perpendicular to the $x$-axis,intersecting it at point $C(k, 0)$.From the equation of the parabola,for $x=k$,$y^{2}=4ak$,so $y=\pm 2\sqrt{ak}$.Thus,the coordinates of $A$ and $B$ are $(k, 2\sqrt{ak})$ and $(k, -2\sqrt{ak})$.The length of the side $AB = 2\sqrt{ak} - (-2\sqrt{ak}) = 4\sqrt{ak}$.Since $OAB$ is an equilateral triangle,$OA=AB$,so $OA^{2}=AB^{2}$.$OA^{2} = k^{2} + (2\sqrt{ak})^{2} = k^{2} + 4ak$.$AB^{2} = (4\sqrt{ak})^{2} = 16ak$.Equating them: $k^{2} + 4ak = 16ak \Rightarrow k^{2} = 12ak$.Since $k 
eq 0$,we have $k = 12a$.The side length $AB = 4\sqrt{a(12a)} = 4\sqrt{12a^{2}} = 4(2\sqrt{3}a) = 8\sqrt{3}a$.

$A. \ 2x-5=0$	$I. \ \text{Vertex}$
$B. \ (\frac{3}{2}, -1)$	$II. \ \text{Focus}$
$C. \ y+1=0$	$III. \ \text{Equation of directrix}$
$D. \ (2, -1)$	$IV. \ \text{Equation of the axis}$
	$V. \ \text{Equation of the Latus rectum}$

An equilateral triangle is inscribed in the parabola $y^{2}=4ax$,where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

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