Let PQ be a focal chord of the parabola y^2=3 | vedclass

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(A) For a parabola $y^2=4ax$,here $4a=36$,so $a=9$. The length of a focal chord with parameter $t$ is $a(t+1/t)^2 = 100$. $9(t+1/t)^2 = 100 \implies (

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$(-3, 43)$

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(A) For a parabola $y^2=4ax$,here $4a=36$,so $a=9$. The length of a focal chord with parameter $t$ is $a(t+1/t)^2 = 100$. $9(t+1/t)^2 = 100 \implies (t+1/t)^2 = 100/9 \implies t+1/t = 10/3$ (since the angle is acute,$t>0$). Solving $t^2 - (10/3)t + 1 = 0$,we get $3t^2 - 10t + 3 = 0$,so $(3t-1)(t-3)=0$. Thus $t=3$ or $t=1/3$. Since the ordinate of $P$ is positive,$P$ corresponds to $t=3$,so $P = (at^2, 2at) = (9 	imes 9, 2 	imes 9 	imes 3) = (81, 54)$. Then $Q$ corresponds to $t=1/3$,so $Q = (9 	imes (1/9), 2 	imes 9 	imes (1/3)) = (1, 6)$. Point $M$ divides $PQ$ in ratio $3:1$,so $M = \frac{3Q+1P}{3+1} = \frac{3(1, 6) + (81, 54)}{4} = \frac{(3+81, 18+54)}{4} = (21, 18)$. The slope of $PQ$ is $m_{PQ} = \frac{54-6}{81-1} = \frac{48}{80} = \frac{3}{5}$. The slope of the line perpendicular to $PQ$ is $m_{\perp} = -5/3$. The equation of the line passing through $M(21, 18)$ with slope $-5/3$ is $(y-18) = -5/3(x-21) \implies 3y-54 = -5x+105 \implies 5x+3y = 159$. Checking the options: $A: 5(-3)+3(43) = -15+129 = 114 
eq 159$. $B: 5(-6)+3(45) = -30+135 = 105 
eq 159$. $C: 5(3)+3(33) = 15+99 = 114 
eq 159$. $D: 5(6)+3(29) = 30+87 = 117 
eq 159$. Given the discrepancy,re-evaluating $M$ with $P(81, 54)$ and $Q(1, 6)$: $M = (21, 18)$. The line is $5x+3y=159$. None of the points lie on this line,suggesting a potential error in the original problem's options or coordinates.

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