The locus of the centroid of an equilateral triangle inscribed in the parabola $y^{2} = 4 a x$ is .

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$9 y^{2} = 2 a (x - 6 a)$

$9 x^{2} = 8 a (y - 4 a)$

$9 x^{2} = 4 a (y + 8 a)$

$9 y^{2} = 4 a (x - 8 a)$

answer is D.

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Detailed Solution

Let P, Q, and R be three points on the parabola.

Let the co-ordinates of the centroid be $G (α, β)$

Clearly $α = \frac{a (t_{1}^{2} + t_{2}^{2} + t_{3}^{2})}{3}$

and $β = \frac{2 a (t_{1} + t_{2} + t_{3})}{3}$

Now,

Slope of $P Q = \frac{2 a (t_{1} - t_{2})}{a (t_{1}^{2} - t_{2}^{2})} = \frac{2}{(t_{1} + t_{2})}$

and the slope of $P R = \frac{2 a (t_{1} - t_{3})}{a (t_{1}^{2} - t_{3}^{2})} = \frac{2}{(t_{1} + t_{3})}$

Clearly $∠ Q P R = 60^{\circ}$

$\tan (60^{\circ}) = \frac{\frac{2}{t_{1} + t_{2}} - \frac{2}{t_{1} + t_{3}}}{1 + \frac{2}{t_{1} + t_{2}} \cdot \frac{2}{t_{1} + t_{3}}}$

$\begin{matrix} \sqrt{3} = \frac{2 (t_{3} - t_{2})}{(t_{1} + t_{2}) (t_{2} + t_{3}) + 4} \\ \sqrt{3} [(t_{1} + t_{2}) (t_{2} + t_{3}) + 4] = 2 (t_{2} - t_{3}) \end{matrix}$

Similarly $∠ Q = 60^{\circ} and ∠ R = 60^{\circ}$

Thus, $\sqrt{3} [(t_{1} + t_{2}) (t_{1} + t_{3}) + 4] = 2 (t_{3} - t_{1})$

$\begin{array}{l} 3 (t_{1} t_{2} + t_{2} t_{3} + t_{3} t_{1}) + (t_{1}^{2} + t_{2}^{2} + t_{3}^{2} + 12) = 0 \\ \Rightarrow 3 (t_{1} t_{2} + t_{2} t_{3} + t_{3} t_{1}) + \frac{3 α}{a} + 12 = 0 \\ \Rightarrow (t_{1} t_{2} + t_{2} t_{3} + t_{3} t_{1}) + \frac{α}{a} + 4 = 0 \\ \Rightarrow a (t_{1} t_{2} + t_{2} t_{3} + t_{3} t_{1}) + α + 4 a = 0 \\ \Rightarrow 2 a (t_{1} t_{2} + t_{2} t_{3} + t_{3} t_{1}) + 2 α + 8 a = 0 \\ \Rightarrow a \{{(t_{1} + t_{2} + t_{3})}^{2} - (t_{1}^{2} + t_{2}^{2} + t_{3}^{2})\} + 2 α + 8 a = 0 \end{array}$

$\begin{array}{l} \Rightarrow a \{\frac{9 β^{2}}{4 a^{2}} - \frac{3 α}{a}\} + 2 α + 8 a = 0 \\ \Rightarrow \frac{9 β^{2}}{4 a} - 3 α + 2 α + 8 a = 0 \end{array}$