Let $\mathrm{S}$ be the focus of the parabola $\mathrm{y}^{2}=8 \mathrm{x}$ and $\mathrm{PQ}$ be the common…
Let $\mathrm{S}$ be the focus of the parabola $\mathrm{y}^{2}=8 \mathrm{x}$ and $\mathrm{PQ}$ be the common chord of the circle $\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}-$ $4 y=0$ and the given parabola. The area of $\Delta P Q S$ is
$4 \mathrm{sq}$ units
$3 \mathrm{sq}$ units
2 sq units
$8 s q$ units
Solution
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[\because t^{2}+t+4 \neq 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 of } \Delta \mathrm{PQS} &=\frac{1}{2} \times \mathrm{QS} \times \mathrm{SP}=\frac{1}{2} \times 2 \times 4 \\
&=4 \mathrm{sq} . \text { units. }
\end{aligned}
$