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(C) The given parabola is $y^2 = 8(x-3)$. Comparing this with $(y-k)^2 = 4a(x-h)$,we get $h=3, k=0$,and $4a=8$,so $a=2$.The focus $S$ is $(h+a, k) = (

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

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(C) The given parabola is $y^2 = 8(x-3)$. Comparing this with $(y-k)^2 = 4a(x-h)$,we get $h=3, k=0$,and $4a=8$,so $a=2$.The focus $S$ is $(h+a, k) = (3+2, 0) = (5, 0)$.The directrix is $x = h-a = 3-2 = 1$. Wait,the directrix is $x = h-a = 3-2 = 1$. Let's re-evaluate.Actually,for $y^2 = 8(x-3)$,the vertex is $(3,0)$ and $a=2$. The directrix is $x = 3-2 = 1$.Let $P$ be $(x, y)$. Since $P$ is on the parabola,$PS = PM$,where $PM$ is the distance to the directrix $x=1$.$PS = PM = |x-1|$.In an equilateral triangle $	riangle SPM$,all sides are equal. Thus $SP = PM = SM = |x-1|$.Since $P$ is on the parabola,$SP = PM$ is always true by definition.For $	riangle SPM$ to be equilateral,we must have $SP = SM$.Let $P = (3+2t^2, 4t)$. Then $S = (5, 0)$.$SP = \sqrt{(3+2t^2-5)^2 + (4t-0)^2} = \sqrt{(2t^2-2)^2 + 16t^2} = \sqrt{4t^4 - 8t^2 + 4 + 16t^2} = \sqrt{4t^4 + 8t^2 + 4} = \sqrt{4(t^2+1)^2} = 2(t^2+1)$.The directrix is $x=1$. The foot of the perpendicular $M$ from $P(3+2t^2, 4t)$ to $x=1$ is $M(1, 4t)$.$SM = \sqrt{(5-1)^2 + (0-4t)^2} = \sqrt{4^2 + 16t^2} = \sqrt{16 + 16t^2} = 4\sqrt{1+t^2}$.Equating $SP = SM$:$2(t^2+1) = 4\sqrt{1+t^2}$.Squaring both sides: $4(t^2+1)^2 = 16(1+t^2)$.$(t^2+1)^2 = 4(t^2+1)$.Since $t^2+1 
eq 0$,we have $t^2+1 = 4$,so $t^2 = 3$,$t = \sqrt{3}$ (taking positive $y$ coordinate).Then $P = (3+2(3), 4\sqrt{3}) = (9, 4\sqrt{3})$.

Point	Slope of the normal
$A. (7, 2)$	$1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$	$2. -8$
$C. (1, -1)$	$3. 4$
$D. (3, \sqrt{2})$	$4. 0$
	$5. -2\sqrt{2}$

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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