Question

Let $S$ be the focus of the parabola $y^2=8x$ and let $PQ$ be the common chord of the circle $x^2+y^2-2x-4y=0$ and the given parabola. The area of the triangle $PQS$ is.

Answer 1

Given: '$S$' be gthe focus of the parabola $y^2=8x$$\therefore$ comparing given parabola with standard parabola $y^2=4ax$.

$\therefore 4a=8$ $a=2$

$\therefore$ focus $S\equiv (a,0)\equiv (2,0)$

Now, $PQ$ is the common chord of circle & parabola

$\therefore$ point $P$ must satisfy equation of parabola & circle.

as any point on parabola is $P\equiv (a{t}^2,2at)$

$\therefore P\equiv (2t^2,4t)......(\because a=2)$

Equation of circle is $x^2+y^2-2x-4y=0$

substituting, C-ordinates of $P$ in rquation of circle.

i.e. $x=2t^2\&y=4t$

$(2t^2{)}^2+(4t{)}^2+2\left(2t^2\right)-4\left(4t\right)=0$

$4t^4+16t^2-4t^2-16t=0$

$4t(t^3+4t-t-4)=0$

$4t(t^3+3t-4)=0$

$\therefore t=0$ and / or $t^3+3t-4=0$ $t=1$

Now, for $t=0,P\equiv (0,0)$

for $t=1,Q\equiv (2,4)$

[ $\because$ Parabola & circle has common chord $\therefore$ one point is $P$ & another is $Q$]

To find Area of $\Delta PQS$,

We have $P\equiv (0,0),Q\equiv (2,4),S\equiv (2,0)$

Method I :

Area $=\frac{1}{2}\times 2\times 4$

$A\left(\Delta PQS\right)=4units$ [Ref. image]

Method II :

Area $=\left|\frac{Px(Qy-Sy)+Qx(Sy-Py)+Sx(Py-Qy)}{2}\right|$

$=\left|\frac{0(4-0)+2(0-0)+2(0-4)}{4}\right|$

$=\left|\frac{0+0-8}{2}\right|$

$Area=4units$