Let P(4, 4sqrt{3}) be a point on the parabola | vedclass

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(C) Given $P(4, 4\sqrt{3})$ lies on $y^2 = 4ax$.Substituting the coordinates of $P$ into the equation: $(4\sqrt{3})^2 = 4a(4) \Rightarrow 48 = 16a \Ri

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$\frac{343\sqrt{3}}{8}$

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(C) Given $P(4, 4\sqrt{3})$ lies on $y^2 = 4ax$.Substituting the coordinates of $P$ into the equation: $(4\sqrt{3})^2 = 4a(4) \Rightarrow 48 = 16a \Rightarrow a = 3$.The equation of the parabola is $y^2 = 12x$. The focus $S$ is $(a, 0) = (3, 0)$.Let the parameter of $P$ be $t_1$. Since $P = (at_1^2, 2at_1) = (3t_1^2, 6t_1) = (4, 4\sqrt{3})$,we have $6t_1 = 4\sqrt{3} \Rightarrow t_1 = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}$.Since $PQ$ is a focal chord,$t_1 t_2 = -1$,so $t_2 = -\frac{\sqrt{3}}{2}$.The coordinates of $Q$ are $(at_2^2, 2at_2) = (3(-\frac{\sqrt{3}}{2})^2, 2(3)(-\frac{\sqrt{3}}{2})) = (3 \cdot \frac{3}{4}, -3\sqrt{3}) = (\frac{9}{4}, -3\sqrt{3})$.The directrix is $x = -a = -3$.The perpendicular distance from $P(4, 4\sqrt{3})$ to $x = -3$ is $PM = 4 - (-3) = 7$.The perpendicular distance from $Q(\frac{9}{4}, -3\sqrt{3})$ to $x = -3$ is $QN = \frac{9}{4} - (-3) = \frac{21}{4}$.The quadrilateral $PQMN$ is a trapezium with parallel sides $PM$ and $QN$ and height $MN$. The length $MN$ is the difference in $y$-coordinates: $MN = |4\sqrt{3} - (-3\sqrt{3})| = 7\sqrt{3}$.Area $= \frac{1}{2} 	imes (PM + QN) 	imes MN = \frac{1}{2} 	imes (7 + \frac{21}{4}) 	imes 7\sqrt{3} = \frac{1}{2} 	imes \frac{49}{4} 	imes 7\sqrt{3} = \frac{343\sqrt{3}}{8}$.

Let $P(4, 4\sqrt{3})$ be a point on the parabola $y^2 = 4ax$ and $PQ$ be a focal chord of the parabola. If $M$ and $N$ are the feet of the perpendiculars drawn from $P$ and $Q$ respectively on the directrix of the parabola,then the area of the quadrilateral $PQMN$ is equal to:

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