Step-by-step Solution

1. Parametric representation of the parabola

   The given parabola is ${y^2} = 2x$. Comparing this with the standard form ${y^2} = 4ax$, we have $4a = 2$, which gives $a = {1 \over 2}$. The parametric coordinates of any point on this parabola can be written as $(at^2, 2at)$, which is $(\frac{1}{2}t^2, t)$.

   Let the coordinates of the two distinct points $P$ and $Q$ be: $P = (\frac{1}{2}t_1^2, t_1)$ $Q = (\frac{1}{2}t_2^2, t_2)$

   The vertex of the parabola is $O(0, 0)$. Since $P$ lies in the first quadrant, its coordinates are positive. This means $x_1 > 0$ and $y_1 > 0$. In parametric form, this implies $t_1 > 0$.

2. Condition of the circle

   A circle with $PQ$ as its diameter passes through the vertex $O(0, 0)$. This means that the angle subtended by the diameter $PQ$ at any point on the circumference is $90^\circ$. Since $O$ is on the circle, we must have $\angle POQ = 90^\circ$. This implies that the lines $OP$ and $OQ$ are perpendicular.

3. Applying the perpendicularity condition

   The slope of the line segment $OP$ is $m_{OP} = \frac{t_1 - 0}{\frac{1}{2}t_1^2 - 0} = \frac{t_1}{\frac{1}{2}t_1^2} = \frac{2}{t_1}$ (since $P$ is not the vertex, $t_1 \neq 0$). The slope of the line segment $OQ$ is $m_{OQ} = \frac{t_2 - 0}{\frac{1}{2}t_2^2 - 0} = \frac{2}{t_2}$ (since $Q$ is not the vertex, $t_2 \neq 0$).

   For perpendicular lines, the product of their slopes is $-1$: $m_{OP} \cdot m_{OQ} = -1$ $\left( \frac{2}{t_1} \right) \left( \frac{2}{t_2} \right) = -1$ $\frac{4}{t_1 t_2} = -1$ $t_1 t_2 = -4$

4. Using the area of the triangle

   The area of triangle $\Delta OPQ$ is given as $3\sqrt{2}$. Since $\angle POQ = 90^\circ$, the area can be calculated as $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} |OP| \cdot |OQ|$.

   Alternatively, we can use the determinant formula for the area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$: Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ For $\Delta OPQ$ with $O(0,0)$, $P(\frac{1}{2}t_1^2, t_1)$, $Q(\frac{1}{2}t_2^2, t_2)$: Area $= \frac{1}{2} |(\frac{1}{2}t_1^2)(t_2) - (\frac{1}{2}t_2^2)(t_1)|$ $3\sqrt{2} = \frac{1}{2} |\frac{1}{2}t_1 t_2 (t_1 - t_2)|$ $3\sqrt{2} = \frac{1}{4} |t_1 t_2 (t_1 - t_2)|$

5. Solving for the parameters

   Substitute $t_1 t_2 = -4$ into the area equation: $3\sqrt{2} = \frac{1}{4} |-4 (t_1 - t_2)|$ $3\sqrt{2} = |-(t_1 - t_2)| = |t_1 - t_2|$ From $t_1 t_2 = -4$, we have $t_2 = -\frac{4}{t_1}$. Substitute this into the equation above: $3\sqrt{2} = |t_1 - (-\frac{4}{t_1})| = |t_1 + \frac{4}{t_1}|$ Since $P$ is in the first quadrant, $t_1 > 0$. Therefore, $t_1 + \frac{4}{t_1}$ is positive. We can remove the absolute value signs. $t_1 + \frac{4}{t_1} = 3\sqrt{2}$ Multiplying by $t_1$ (since $t_1 \neq 0$): $t_1^2 + 4 = 3\sqrt{2} t_1$ $t_1^2 - 3\sqrt{2} t_1 + 4 = 0$ This is a quadratic equation for $t_1$. We solve it using the quadratic formula, $t_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $t_1 = \frac{-(-3\sqrt{2}) \pm \sqrt{(-3\sqrt{2})^2 - 4(1)(4)}}{2(1)}$ $t_1 = \frac{3\sqrt{2} \pm \sqrt{18 - 16}}{2}$ $t_1 = \frac{3\sqrt{2} \pm \sqrt{2}}{2}$ This gives two possible values for $t_1$: $t_1 = \frac{3\sqrt{2} + \sqrt{2}}{2} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$ or $t_1 = \frac{3\sqrt{2} - \sqrt{2}}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2}$

6. Finding the coordinates of P

   The coordinates of $P$ are $(\frac{1}{2}t_1^2, t_1)$.

   Case 1: $t_1 = 2\sqrt{2}$ $x_1 = \frac{1}{2}(2\sqrt{2})^2 = \frac{1}{2}(8) = 4$ $y_1 = t_1 = 2\sqrt{2}$ So, one possible point for $P$ is $(4, 2\sqrt{2})$.

   Case 2: $t_1 = \sqrt{2}$ $x_1 = \frac{1}{2}(\sqrt{2})^2 = \frac{1}{2}(2) = 1$ $y_1 = t_1 = \sqrt{2}$ So, another possible point for $P$ is $(1, \sqrt{2})$.

7. Conclusion

   The possible coordinates for $P$ are $(4, 2\sqrt{2})$ and $(1, \sqrt{2})$. These correspond to options A and D.

Evaluation of Options

• A: $(4, 2\sqrt{2})$: This matches one of our derived coordinates. Correct.
• B: $(9, 3\sqrt{2})$: This point is on the parabola since $(3\sqrt{2})^2 = 18 = 2(9)$. Here $t_1 = 3\sqrt{2}$. Then $t_1 + 4/t_1 = 3\sqrt{2} + 4/(3\sqrt{2}) = (18+4)/(3\sqrt{2}) = 22/(3\sqrt{2}) \neq 3\sqrt{2}$. Incorrect.
• C: $(\frac{1}{4}, \frac{1}{\sqrt{2}})$: This point is on the parabola since $(1/\sqrt{2})^2 = 1/2 = 2(1/4)$. Here $t_1 = 1/\sqrt{2}$. Then $t_1 + 4/t_1 = 1/\sqrt{2} + 4/(1/\sqrt{2}) = 1/\sqrt{2} + 4\sqrt{2} = (1+8)/\sqrt{2} = 9/\sqrt{2} \neq 3\sqrt{2}$. Incorrect.
• D: $(1, \sqrt{2})$: This matches one of our derived coordinates. Correct.