Question

An equilateral triangle is inscribed in the parabola $y^2=4ax$ so that one angular point of the triangle is at the vertex of the parabola. Find the length of each side of the triangle.

Answer 1

Let $△OAB$ be an equilateral triangle inscribed in the parabola $y^2=4ax.$

Let $OA=OB=AB=1.$

Let AB cut the x-axis at M.

then, $\angle AOM=\angle BOM={30}^{\circ }.$

$\therefore \frac{OM}{OA}=cos{30}^{\circ }\Rightarrow OM=lcos{30}^{\circ }=\frac{l\sqrt{3}}{2},$

and $\frac{AM}{OA}=sin{30}^{\circ }\Rightarrow AM=lsin{30}^{\circ }=\frac{1}{2}.$

$\therefore$ the coordinates of A are $(\frac{l\sqrt{3}}{2},\frac{1}{2}).$

Since A lies on the parabola $y^2=4ax,$ we have

$\frac{l^2}{4}=4a\times \frac{l\sqrt{3}}{2}\Rightarrow l=8a\sqrt{3}.$

Hence, the length of each side of the triangle is $8a\sqrt{3}$ units.