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(B) Let $a$ be the length of the side of the equilateral triangle.Since the triangle is symmetric about the $x$-axis,one vertex is at the origin $(0,

Vedclass · Accepted Answer

$16 \sqrt{3} 	ext{ units}$

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(B) Let $a$ be the length of the side of the equilateral triangle.Since the triangle is symmetric about the $x$-axis,one vertex is at the origin $(0, 0)$.The other two vertices are at $(x, y)$ and $(x, -y)$.For an equilateral triangle with side $a$,the height is $\frac{\sqrt{3}}{2}a$ and the vertical distance from the $x$-axis to the vertices is $\frac{a}{2}$.Thus,the coordinates of the vertex on the parabola are $\left(\frac{\sqrt{3}}{2}a, \frac{a}{2}ight)$.Since this point lies on the parabola $y^2 = 8x$,we substitute the coordinates:$\left(\frac{a}{2}ight)^2 = 8 \left(\frac{\sqrt{3}}{2}aight)$$\frac{a^2}{4} = 4\sqrt{3}a$Since $a 
eq 0$,we divide by $a$:$\frac{a}{4} = 4\sqrt{3}$$a = 16\sqrt{3} 	ext{ units}$.

An equilateral triangle is inscribed in the parabola $y^2 = 8x$,with one of its vertices at the vertex of the parabola. Then,the length of the side of that triangle is

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