Let P and Q be distinct points on the parabol | vedclass

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(D) The equation of the parabola is $y^2 = 2x$,so $4a = 2$,which gives $a = \frac{1}{2}$.Let $P = (\frac{t_1^2}{2}, t_1)$ and $Q = (\frac{t_2^2}{2}, t

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$(A, D)$

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(D) The equation of the parabola is $y^2 = 2x$,so $4a = 2$,which gives $a = \frac{1}{2}$.Let $P = (\frac{t_1^2}{2}, t_1)$ and $Q = (\frac{t_2^2}{2}, t_2)$ be points on the parabola.Since the circle with diameter $PQ$ passes through the origin $O(0,0)$,the vectors $\vec{OP}$ and $\vec{OQ}$ are perpendicular,so $\vec{OP} \cdot \vec{OQ} = 0$.$(\frac{t_1^2}{2})(\frac{t_2^2}{2}) + t_1 t_2 = 0 \Rightarrow t_1 t_2 (\frac{t_1 t_2}{4} + 1) = 0$.Since $P$ and $Q$ are distinct,$t_1 t_2 = -4$.Let $t_1 = t$,then $t_2 = -\frac{4}{t}$.The coordinates are $P = (\frac{t^2}{2}, t)$ and $Q = (\frac{8}{t^2}, -\frac{4}{t})$.The area of $\Delta OPQ = \frac{1}{2} |x_P y_Q - x_Q y_P| = \frac{1}{2} |(\frac{t^2}{2})(-\frac{4}{t}) - (\frac{8}{t^2})(t)| = \frac{1}{2} |-2t - \frac{8}{t}| = |t + \frac{4}{t}|$.Given area is $3\sqrt{2}$,so $|t + \frac{4}{t}| = 3\sqrt{2}$.Since $P$ is in the first quadrant,$t > 0$,so $t^2 - 3\sqrt{2}t + 4 = 0$.Solving for $t$: $t = \frac{3\sqrt{2} \pm \sqrt{18 - 16}}{2} = \frac{3\sqrt{2} \pm \sqrt{2}}{2}$.$t_1 = 2\sqrt{2}$ and $t_2 = \sqrt{2}$.For $t = 2\sqrt{2}$,$P = (\frac{(2\sqrt{2})^2}{2}, 2\sqrt{2}) = (4, 2\sqrt{2})$.For $t = \sqrt{2}$,$P = (\frac{(\sqrt{2})^2}{2}, \sqrt{2}) = (1, \sqrt{2})$.Thus,the coordinates of $P$ are $(4, 2\sqrt{2})$ and $(1, \sqrt{2})$,which corresponds to option $(D)$.

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