Explanation

(b)
Since, tangent at $P$ and $Q$ on the parabola meet in $T$
If the coordinates of $P$ and $Q$ are $(at_{1}^{2},\ 2at_{1})$ and $(at_{2}^{2},\ 2at_{2})$ respectively, then coordinates of $T$ are $\{ at_{1}t_{2},\ a(t_{1} + t_{2})\}$
$\therefore SP = a(1 + t_{1}^{2})$
$SP = a(1 + t_{2}^{2})$
${ST}^{2} = a^{2}{(1 - t_{1}t_{2})}^{2} + a^{2}{(t_{1} + t_{2})}^{2}$
$= a^{2}(1 + t_{1}^{2} + t_{2}^{2} + t_{1}^{2}t_{2}^{2})$
$= a\left( 1 + t_{1}^{2} \right)a\left( 1 + t_{2}^{2} \right) = SP \bullet SQ$
Thus, $SP,\ ST,\ SQ$ are in GP