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(C) Given that the $	riangle SPM$ is equilateral.For the parabola $y^2=8(x-3)$,we have $4a=8$,so $a=2$.The vertex is $(3, 0)$ and the focus $S$ is $(

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$(9, 4 \sqrt{3})$

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(C) Given that the $	riangle SPM$ is equilateral.For the parabola $y^2=8(x-3)$,we have $4a=8$,so $a=2$.The vertex is $(3, 0)$ and the focus $S$ is $(3+a, 0) = (5, 0)$.The directrix is $x = 3-a = 3-2 = 1$. Wait,the directrix is $x = 3-2 = 1$. Let us re-evaluate.For $y^2=4a(x-h)$,directrix is $x=h-a$. Here $h=3, a=2$,so $x=3-2=1$.Let $P$ be $(x, y)$. Since $P$ is on the parabola,$PS = PM$,where $PM$ is the distance to the directrix $x=1$.$PS = \sqrt{(x-5)^2 + y^2} = \sqrt{(x-5)^2 + 8(x-3)} = \sqrt{x^2-10x+25+8x-24} = \sqrt{x^2-2x+1} = |x-1|$.$PM = |x-1|$. This confirms the definition of a parabola.In $	riangle SPM$,$PS=PM$. For it to be equilateral,$PS=PM=SM$.$SM$ is the distance from $S(5, 0)$ to $M(1, y)$. $SM = \sqrt{(5-1)^2 + (0-y)^2} = \sqrt{16+y^2}$.Since $PS = |x-1|$,we have $PS^2 = (x-1)^2$.Also $y^2 = 8(x-3)$,so $PS^2 = (x-1)^2 = x^2-2x+1$.Equating $PS^2 = SM^2$: $(x-1)^2 = 16+y^2$.$(x-1)^2 = 16 + 8(x-3) = 16 + 8x - 24 = 8x - 8$.$x^2 - 2x + 1 = 8x - 8 \Rightarrow x^2 - 10x + 9 = 0$.$(x-9)(x-1) = 0$. Since $x=1$ is the directrix,$x=9$.If $x=9$,$y^2 = 8(9-3) = 8(6) = 48$,so $y = \pm 4\sqrt{3}$.Thus $P = (9, 4\sqrt{3})$ or $(9, -4\sqrt{3})$.

Let $M$ be the foot of the perpendicular from a point $P$ on the parabola $y^2=8(x-3)$ onto its directrix and let $S$ be the focus of the parabola. If $\triangle SPM$ is an equilateral triangle,then $P$ is equal to

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