Let S be the focus of the parabola ^ 2 =8 x and PQ be the common...

The parametric equations of the parabola $y^{2}=8 x$ are $x=2 t^{2} and y=4 t and the given equation of circle is$x^{2}+y^{2}-2 x-4 y=0$On putting$x=2 t^{2} and y=4$t in circle we get$$\begin{array}{l}4 t^{4}+16 t^{2}-4 t^{2}-16 t=0 \\ \Rightarrow 4 t^{2}+12 t^{2}-16 t=0 \\ \Rightarrow 4 t \left(t^{3}+3 t-4\right)=0 \\ \Rightarrow t(t-1) \left(t^{2}+t+4\right)=0 \\ \Rightarrow t=0, t=1 \\ \quad \left[\becauset^{2}+t+4 ≠ 0\right] \end{array}$$Thus$ the coordinates of points of intersection of the circle and the parabola are $\mathrm{Q}(0,0) and \mathrm{P}(2$,4) .Clearly these are diametrically opposite points on the $circle. The coordinates of the focus \mathrm{S} of the parabola are (2,0) which lies on the circle.$$\begin{aligned} \therefore \text{ Area\text{ of } } \Delta \mathrm{PQS}&=\frac{1}{2} \times \mathrm{QS} \times \mathrm{SP}=\frac{1}{2} \times2 \times4 \\ &=4 \mathrm{sq} . \text{ units. } \end{aligned}$$