Question: Let $S_1$ be the focus of the parabola $$y^2 = 8x$$ and $PQ$ 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?
I circumvented this problem knowing that the circle touches the origin as it has no constant value and so does the standard parabola. I assumed (accidentally and also correctly) that the chord was the diameter, knowing the centre was $(1,2)$ and I found the other vertex as $(2,4)$ and solved the question getting the correct answer.
Is there perhaps a generalised method to find the equation of the parabola and the circle? I know that the chord of a parabola is given by the formula $y(t_1+1_2)=2at_1t_1$, but I haven't figured how to use it to find actual equations of chord apart the formula
To find the common chord of the parabola and the circle, we must solve their equations simultaneously. Substituting $x = \frac{y^2}{8}$ in the equation for the circle: $$ \frac{y^4}{64} + y^2- 2\frac{y^2}{8} - 4y = 0 \qquad \implies \qquad y^4 + 48y^2 - 256y = 0$$ $\therefore$ Either $y = 0, \implies x = 0,$ or $y^3 -48y -256 = 0.$ By rational root theorem, $y=4$ satisfies the cubic equation. $y = 4 \implies x = 2$. Thus $(0,0)$ and $(2,4)$ are the points of intersection. Let $P = (0,0)$ and $Q = (2,4)$. We know that $S_1 = (2,0)$. Now we can use Heron's formula to calculate the area.