Mathematics - Parabola Question with Solution | TestHub

Given parabola $y^2=8x$ and circle $x^2+y^2-2x-4y=0$ pass through the origin $\therefore$ One end of common chord PQ is origin. Say $\mathrm{P}(0,0)$ Let $\mathrm{Q}$ be the point $\left(2t^2,4t\right)$, then it will satisfy the equation of circle. $\therefore 4t^4+16t^2-4t^2-16t=0$ $\Rightarrow t^4+3t^2-4t=0\Rightarrow t\left(t^3+3t-4\right)=0$ $\Rightarrow t(t-1)\left(t^2+t-4\right)=0\Rightarrow t=0$ or 1 For $t=0$, we get point $P$, therefore $t=1$ gives point $Q$ as $(2,4)$. We also observe here that $\mathrm{P}(0,0)$ and $\mathrm{Q}(2,4)$ are end points of diameter of the given circle and focus of the parabola is the point $S(2,0)$. $\therefore$ area $(\Delta \mathrm{PQS})=\frac{1}{2}\times PS\times QS=\frac{1}{2}\times 2\times 4=4$