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

$y^2=8x\dots \left(i\right)$
$x^2+y^2-2x-4y=0\dots \left(ii\right)$

From $\left(i\right)$ and $\left(ii\right),$ we get
$x^2+8x-2x-4\times 2\sqrt{2}\sqrt{x}=0$
$\Rightarrow x^2+6x-8\sqrt{2x}=0$
$\Rightarrow \sqrt{x}(x^{3/2}+6x^{1/2}-8\sqrt{2})=0$

Let $\sqrt{x}=t,$ then
$t^3+6t-8\sqrt{2}=0\Rightarrow (t-\sqrt{2})(t^2-\sqrt{2}t+4)=0$
$\Rightarrow t=\sqrt{2}\Rightarrow x=2,y=4$

$\therefore P\equiv (2,4),Q\equiv (0,0)$ and $S(2,0).$

Area of $△PQS$
$=\frac{1}{2}\times 2\times 4=4$