Question

Let ( mathrm{OAB} ) be the equilateral triangle inscribed in parabola ( y^{2}=4 a x )
Let ( A B ) intersect the ( x ) -axis at point ( C ).
Let ( O C=k )
From the equation of the given parabola, we have ( y^{2}=4 a k Rightarrow y=pm 2 sqrt{a k} )
The respective coordinates of points ( A ) and ( B ) are ( (k, 2 sqrt{a k}), ) and ( (k,-2 sqrt{a k}) )
( A B=C A+C B=2 sqrt{a k}+2 sqrt{a k}=4 sqrt{a k} )
since ( mathrm{OAB} ) is an equilateral triangle, ( mathrm{OA}^{2}=mathrm{AB}^{2} )
( therefore k^{2}+(2 sqrt{a k})^{2}=(4 sqrt{a k})^{2} )
( Rightarrow k^{2}+4 a k=16 a k )
( Rightarrow k^{2}=12 a k )
( Rightarrow k=12 a )
( therefore A B=4 sqrt{a k}=4 sqrt{a times 12 a}=4 sqrt{12 a^{2}}=8 sqrt{3} a )
Thus, the side of the equilateral triangle inscribed in parabola ( y^{2}=4 a x ) is ( 8 sqrt{3} a ).

# 1 | MIU POJLJ Clubu Uy tic Ild, 8. An equilateral triangle is inscribed in the parabola y2 = 4 ax, where one vertex is at the vertex of the parabola. Find the length of the side of che triangle.

Solution