If the focal distance on a point to a parabola $y^2=12x$ is $6$, find the co-ordinates of that point. 

Let the point be $(x, y)$

The given equation is

$\begin{array}{l}{y}^2=12x\\ \Rightarrow 4a=12\\ \Rightarrow a=12/4=3\end{array}$

Given focal distance = 6 

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x+a=6\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x+3=6\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=6-3=3\end{array}$

When $x=3, y^2=12\times 3=36$

$\Rightarrow y=6$

Hence, the co-ordinates of the points are $\left(3, 6\right)$ and  $\left(3, -6\right)$.