Q.Find the Cartesian co-ordinates of the points on the parabola $y^2 = 12x$ whose parameter is 2.

Rather than working with the two coupled variables $x,y$ satisfying one equation, it is often far simpler to describe every point on a parabola using a single free variable $t$, called the parameter. For the standard parabola $y^2=4ax$, the point $(at^2,\,2at)$ lies on the curve for every real value of $t$ — this can be checked directly by substitution: $(2at)^2=4a^2t^2=4a(at^2)$.

This parametric description, written $P(t)\equiv(at^2,2at)$, is what makes later tangent-line derivations so clean: the parametric-form tangent at the point with parameter $t_1$ turns out to be the strikingly simple linear equation $yt_1=x+at_1^2$, compared with the point-form's $yy_1=2a(x+x_1)$.

A practical tip for finding the parameter of a given point: since $y=2at$ is linear in $t$, solving $t=\dfrac{y}{2a}$ directly from the $y$-coordinate is much safer than using $x=at^2$ (which is quadratic in $t$ and only pins down $t$ up to a sign) — always verify the result against the $x$-coordinate afterward as a check.