On comparing the equation of the parabola
$y^{2}=12 x$ with the standard equation,
$y^{2}=4$ ax, we get $4 a=12$ or $a=3$


Hence, the focus, point $\mathrm{C}$ will be at $(3,0)$ and the extremities of the latus-rectum $\mathrm{AB}$ will be at $(a, 2 a)$ and $(a,-2 a)$. So the coordinates of A and $B$ are $(3,6)$ and $(3,-6)$ respectively. Now we need to find the length ( curve $\mathrm{AOB}$ ) of the parabola. As it is not a straight line so we cannot directly find the length of this curve as we cannot directly apply Pythagorous theorem. Let us consider a small length $\mathrm{ds}$ on the parabola. Using pythagorous theorem for this length,
$\mathrm{ds}=\sqrt{(\mathrm{dx})^{2}+(\mathrm{dy})^{2}}=\sqrt{\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)^{2}+1 \mathrm{dy}}$
$\Rightarrow \mathrm{s}=\int_{-6}^{6}\left[\sqrt{\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)^{2}+1}\right] \cdot \mathrm{dy}$ ...(1)
From $y^{2}=12 x \Rightarrow x=\frac{y^{2}}{12}$
$\frac{d x}{d y}=\frac{2 y}{12}=\frac{y}{6} \quad$ Putting in (1),
$s=\int_{-6}^{6}\left[\sqrt{\left(\frac{y}{6}\right)^{2}+1}\right] d y=2 \int_{0}^{6} \sqrt{\frac{y^{2}+36}{36}} d y$
$=\frac{2}{6} \int_{0}^{6} \sqrt{y^{2}+6^{2}} d y$
Using $\int \sqrt{x^{2}+a^{2}}=\frac{x}{2} \sqrt{a^{2}+x^{2}}$
$+\frac{a^{2}}{2} \log \left(x+\sqrt{a^{2}+x^{2}}\right)+C$
We get $s=\frac{1}{3}\left[\frac{y}{2} \sqrt{6^{2}+y^{2}}+\frac{6^{2}}{2}\right.$
$\left.\log \left(y+\sqrt{6^{2}+y^{2}}\right)+C\right]_{0}^{6}$
$=\frac{1}{3}\left[\frac{6}{2} \sqrt{6^{2}+6^{2}}+18 \log \left(6+\sqrt{6^{2}+6^{2}}\right)\right.$
$\left.\quad+C-0-18 \log \left(0+\sqrt{6^{2}+0}\right)-C\right]$
$=\frac{1}{3}[3.6 \sqrt{2}+18 \log (6+6 \sqrt{2})-18 \log 6]$
$=6 \sqrt{2}+6 \log \frac{6(1+\sqrt{2})}{6}$
$=6[(\sqrt{2}+\log (1+\sqrt{2})]$