Answer:
Option B
Explanation:
Concept Involved Parametric coordinates for $y^{2} = 4ax$ are$ (at^{2},2at).$
Description of Situation As the circle intersects the parabola at P and Q. Thus points P and Q should satisfy the circle.
Sol. $p(2t^{2},4t)$ sholud lie on
$x^{2}+y^{2}-2x-4y=0$
$\Rightarrow$ $4t^{4}+16t^{2}-4t^{2}-16 T-0$
$\Rightarrow$ $4t^{4}+12t^{2}-16t=0$
$\Rightarrow$ $4t{(t^{3}+3t-4)}=0$
$\Rightarrow$ $4t(t-1)(t^{2}+t+4)=0$
$\therefore$ $t=0,1 \Rightarrow P(2,4)$
Thus, are of $\triangle OPs=\frac{1}{2} OS \times PQ$