Areal Parabolic Centroid

Area made by chord joining $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$ of parabola $y^2 = 4ax$ with parabola is

$A = \frac{a^2}{3}(t_2-t_1)^3$

$\boxed{t_2 = t_1 + \sqrt[3]{\frac{3A}{a^2}}}$

Centroid of $\Delta OPQ$ is

$G(h, k) \equiv \bigg(\frac{a}{3}(t_1^2 + t_2^2), \frac{a}{3}(t_1 + t_2)\bigg)$

Putting value of $t_2$

$G(h, k) \equiv \bigg(\frac{a}{3}\bigg(t_1^2 +\bigg(t_1 + \sqrt[3]{\frac{3A}{a^2}}\bigg)^2\bigg), \frac{a}{3}\bigg(t_1 + \bigg(t_1 + \sqrt[3]{\frac{3A}{a^2}}\bigg)\bigg)\bigg)$

Solving and eliminating $t_1$ from $(h, k)$ , we get

$\boxed{k^2 = \frac{8a}{3}\bigg(h - \frac{a\bigg(\sqrt[3]{\frac{3A}{a^2}}\bigg)^2}{6}\bigg)}$

So, locus is

$\boxed{y^2 = \frac{8a}{3}\bigg(x - \frac{a\bigg(\sqrt[3]{\frac{3A}{a^2}}\bigg)^2}{6}\bigg)}$

So, length of latus rectum is $L = \frac{8a}{3}$ and focus is $S(\frac{2a}{3} + \frac{a(\sqrt[3]{\frac{3A}{a^2}})^2}{6}, 0)$

Putting $a = 6$ and for minimum $x$ , $A = 12$

$\color{#3D99F6}{\boxed{L = 16, x \gt 5, y = 0}}$

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So, $a = 16, b = 5, c = 0, a + b + c = 21$