Mathematics - Parabola Question with Solution | TestHub

The parametric equations of the parabola $y^2=8x$ are $x=2t^2$ and $y=4t$ and the given equation of circle is $x^2+y^2-2x-4y=0$ On putting $x=2t^2$ and $y=4$ t in circle we get $$\begin{array}{l}4t^4+16t^2-4t^2-16t=0\\ \Rightarrow 4t^2+12t^2-16t=0\\ \Rightarrow 4t\left(t^3+3t-4\right)=0\\ \Rightarrow t(t-1)\left(t^2+t+4\right)=0\\ \Rightarrow t=0,t=1\\ \left[\because t^2+t+4\ne 0\right]\end{array}$$ 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. $$\begin{array}{rl}\therefore \text{ Area of }\Delta \mathrm{PQS}& =\frac{1}{2}\times \mathrm{QS}\times \mathrm{SP}=\frac{1}{2}\times 2\times 4\\ & =4\mathrm{sq}.\text{ units. }\end{array}$$