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  • The locus of the centroid of equilateral triangles inscribed in the parabola

The locus of the centroid of equilateral triangles inscribed in the parabola \mathrm{y^2=4 a x} is the parabola \mathrm{9 \mathrm{y}^2-\mathrm{kxa}+32 \mathrm{a}^2=0} where \mathrm{\mathrm{k}=}

Option: 1

1


Option: 2

4


Option: 3

8


Option: 4

12


Answers (1)

best_answer

Let vertices of equilateral \mathrm{\triangle P Q R} be \mathrm{P\left(a t_1^2, 2 a t_1\right), Q\left(a t_2^2, 2 a t_2\right)} and \mathrm{R\left(a t_3^2, 2 a t_3\right).}
If coordinates of centroid G is \mathrm{(h, k)}, then
\mathrm{ \frac{\mathrm{h}}{\mathrm{a}}=\frac{\mathrm{t}_1^2+\mathrm{t}_2^2+\mathrm{t}_3^2}{3} }                              .......(1)
and \mathrm{\frac{\mathrm{k}}{2 \mathrm{a}}=\frac{\mathrm{t}_1+\mathrm{t}_2+\mathrm{t}_3}{3}}                           .......(2)
As slope of PQ is \mathrm{\frac{2}{t_1+t_2}} and of QR is \mathrm{\frac{2}{t_2+t_3}}

\mathrm{ \therefore \frac{\frac{2}{t_1+t_2}-\frac{2}{t_2+t_3}}{1+\frac{4}{\left(t_1+t_2\right)\left(t_2+t_3\right)}}=\tan 60^{\circ}=\sqrt{3} }

\mathrm{\Rightarrow 2\left(t_3-t_1\right)=4 \sqrt{3}+\sqrt{3}\left(\sum t_1 t_2+t_2^2\right)}       .......(3)

\mathrm{\text { Similarly } 2\left(t_1-t_2\right)=4 \sqrt{3}+\sqrt{3}\left(\sum t_1 t_2+t_3^2\right) \ldots \ldots}(4)

\mathrm{2\left(t_2-t_3\right)=4 \sqrt{3}+\sqrt{3}\left(\sum t_1 t_2+t_1^2\right)}       

Adding equation (3), (4) and (5) 

\mathrm{ 0=12 \sqrt{3}+3 \sqrt{3} \sum t_1 t_2+\left(t_1^2+t_2^2+t_3^2\right) \sqrt{3} }
So by equation (1) and (2)

\mathrm{ \begin{aligned} & 12+\frac{3}{2}\left(\frac{9 k^2}{4 a^2}-\frac{3 h}{a}\right)+\frac{3 h}{a}=0 \\ & \Rightarrow 9 k^2-4 a h+32 a^2=0 \\ & \therefore \text { locus is } 9 y^2-4 a x+32 a^2=0 . \end{aligned} }

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