Let S be the focus of the parabola y^2=8x and | vedclass

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(D) The focus of the parabola $y^2=8x$ is $S \equiv (2, 0)$.To find the common chord,subtract the equation of the parabola from the equation of the ci

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$4$

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(D) The focus of the parabola $y^2=8x$ is $S \equiv (2, 0)$.To find the common chord,subtract the equation of the parabola from the equation of the circle:$(x^2+y^2-2x-4y) - (y^2-8x) = 0$$x^2+6x-4y = 0$Since $y^2=8x$,we substitute $x = \frac{y^2}{8}$ into the equation of the chord:$(\frac{y^2}{8})^2 + 6(\frac{y^2}{8}) - 4y = 0$$\frac{y^4}{64} + \frac{3y^2}{4} - 4y = 0$$y^4 + 48y^2 - 256y = 0$$y(y^3 + 48y - 256) = 0$One solution is $y=0$,which gives $x=0$. Thus,$P \equiv (0, 0)$.For $y^3 + 48y - 256 = 0$,by inspection,$y=4$ is a root $(64 + 192 - 256 = 0)$.If $y=4$,then $x = \frac{16}{8} = 2$. Thus,$Q \equiv (2, 4)$.The vertices of $	riangle PQS$ are $P(0, 0)$,$Q(2, 4)$,and $S(2, 0)$.Area $= \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$Area $= \frac{1}{2} |0(4-0) + 2(0-0) + 2(0-4)|$Area $= \frac{1}{2} |0 + 0 - 8| = \frac{1}{2} |-8| = 4$.

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:

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