Question

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.

JEE/Engineering Exams

Maths

Solution

121

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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 ).