Let $$x = 2t, y = \frac{t^2}{3}$$ be a conic. Let $$S$$ be the focus and $$B$$ be the point on the axis of the conic such that $$SA \perp BA$$, where $$A$$ is any point on the conic. If $$k$$ is the ordinate of the centroid of the $$\triangle SAB$$, then $$\lim_{t \to 1} k$$ is equal to

We have the parametric curve $$x = 2t,\;y = \frac{t^2}{3}$$ and need to find $$\lim_{t\to 1}k$$ where $$k$$ is the ordinate of the centroid of triangle $$SAB$$.

Since $$x = 2t$$ gives $$t = x/2$$, substituting into $$y = \frac{t^2}{3}$$ yields $$y = \frac{x^2}{12}$$ or equivalently $$x^2 = 12y\,, $$ which is the equation of a parabola with $$4a = 12$$ so that $$a = 3$$. Hence its focus is $$S = (0,3)$$ and its vertex is at the origin.

Any point on this parabola can be represented as $$A = (2t,\;t^2/3)$$, and if we let $$B = (0,b)$$ lie on its axis then the vectors from $$S$$ to $$A$$ and from $$B$$ to $$A$$ become $$\vec{SA} = (2t,\;t^2/3 - 3)$$ and $$\vec{BA} = (2t,\;t^2/3 - b)$$, respectively.

Requiring $$SA$$ to be perpendicular to $$BA$$ means their dot product must vanish, so
$$(2t)(2t) + \Bigl(\tfrac{t^2}{3}-3\Bigr)\Bigl(\tfrac{t^2}{3}-b\Bigr)=0\quad\Longrightarrow\quad 4t^2 + \Bigl(\tfrac{t^2}{3}-3\Bigr)\Bigl(\tfrac{t^2}{3}-b\Bigr)=0\,. $$

At this stage it is convenient to set $$y_A = t^2/3$$, which implies $$t^2 = 3y_A$$, and substitution leads to
$$12y_A + (y_A-3)(y_A-b)=0\quad\Longrightarrow\quad 12y_A + y_A^2 - y_A b -3y_A +3b=0\quad\Longrightarrow\quad b(3-y_A)=-(y_A^2+9y_A)\,. $$

From which it follows that
$$b=\frac{y_A^2+9y_A}{y_A-3}\,. $$

As $$t\to 1$$ we have $$y_A=1/3$$, hence
$$b=\frac{(1/3)^2+9(1/3)}{1/3-3}=\frac{1/9+3}{-8/3}=\frac{28/9}{-8/3}=\frac{28}{9}\times\frac{-3}{8}=-\frac{7}{6}\,. $$

Finally, with $$S=(0,3)$$, $$A=(2,1/3)$$ and $$B=(0,-7/6)$$, the ordinate of the centroid is given by
$$k=\frac{y_S+y_A+y_B}{3}=\frac{3+\tfrac{1}{3}-\tfrac{7}{6}}{3}=\frac{\tfrac{18}{6}+\tfrac{2}{6}-\tfrac{7}{6}}{3}=\frac{13/6}{3}=\frac{13}{18}\,. $$

Option D: $$\dfrac{13}{18}$$.