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(A) The equation of the parabola is $y^2=16x$. The vertex of the parabola is at the origin $O(0,0)$. Let the vertices of the equilateral triangle be $

Vedclass · Accepted Answer

$32 \sqrt{3}$

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(A) The equation of the parabola is $y^2=16x$. The vertex of the parabola is at the origin $O(0,0)$. Let the vertices of the equilateral triangle be $O(0,0)$,$A(4t^2, 8t)$,and $B(4t^2, -8t)$.Since the triangle is equilateral,the angle $\angle AOM = 30^{\circ}$,where $M$ is the projection of $A$ on the $X$-axis.In $	riangle AOM$,$	an 30^{\circ} = \frac{AM}{OM} = \frac{8t}{4t^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 coordinates of point $A$ are $(4(2\sqrt{3})^2, 8(2\sqrt{3})) = (4(12), 16\sqrt{3}) = (48, 16\sqrt{3})$.The length of the side of the triangle is the distance $OA = \sqrt{(48-0)^2 + (16\sqrt{3}-0)^2} = \sqrt{48^2 + 256 	imes 3} = \sqrt{2304 + 768} = \sqrt{3072}$.$\sqrt{3072} = \sqrt{1024 	imes 3} = 32\sqrt{3}$.Thus,the length of the side of the equilateral triangle is $32\sqrt{3}$.

If all the vertices of an equilateral triangle lie on the parabola $y^2=16x$ and one of them coincides with the vertex of that parabola,then the length of the side of that triangle is

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