$P Q$ is a variable focal chord of the parabola $y^{2} = 4 a x$ whose vertex is $A$ . Prove that the locus of the centroid of triangle $A P Q$ is a parabola whose latus rectum is $\frac{4 a}{3}$ .

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Solution

Consider $2$ pts P & Q are parabola as shown. As PQ pass through focus.
Thus, $t_{1} t_{2} = - 1$

Centroid of $Δ P A Q (\frac{a t_{1}^{2} + a t_{2}^{2}}{3}, \frac{2 a t_{1} + 2 a t_{2}}{3})$

So $x = \frac{a}{3} (t_{1}^{2} + t_{2}^{2})$ $y = \frac{2 a}{3} (t_{1} + t_{2})$

$\Rightarrow x = \frac{a}{3} [(t_{1} + t_{2})^{2} - 2 t_{1} t_{2}]$

$\Rightarrow x = \frac{a}{3} {{(\frac{3 y}{2 a})}^{2} - 2 (- 1)}$

$\Rightarrow 3 x = a {\frac{9 y^{2}}{4 a^{2}} + 2}$

$\Rightarrow 3 x = {\frac{9 y^{2} + 8 a^{2}}{4 a}}$

$\Rightarrow 9 y^{2} + 8 a^{2} = 12 a x$

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

$\Rightarrow y^{2} = (\frac{4 a}{9}) (x - 2 a)$

Clearly its latus rectum is $(4 a / 3)$ .

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