Coordinates of Point on Parabola with Given Triangle Area

To find the possible coordinates of point $P$, we proceed with the following steps:

1. Parametric Representation: The given equation of the parabola is: $y^2 = 2x = 4\left(\frac{1}{2}\right)x$ Here, $a = \frac{1}{2}$. Any point on this parabola can be represented in parametric coordinates $(x, y) = \left(at^2, 2at\right) = \left(\frac{t^2}{2}, t\right)$.

Let the coordinates of the distinct points $P$ and $Q$ be: $P = \left(\frac{t_1^2}{2}, t_1\right) \quad \text{and} \quad Q = \left(\frac{t_2^2}{2}, t_2\right)$ Since $P$ lies in the first quadrant, we have $t_1 > 0$.

1. Condition for the Circle Passing through the Vertex $O(0,0)$: Since the circle described on $PQ$ as a diameter passes through the vertex $O(0,0)$, the angle subtended by the diameter $PQ$ at the origin must be a right angle: $\angle POQ = 90^\circ \implies OP \perp OQ$ The slopes of lines $OP$ and $OQ$ are: $m_{OP} = \frac{t_1 - 0}{\frac{t_1^2}{2} - 0} = \frac{2}{t_1}$ $m_{OQ} = \frac{t_2 - 0}{\frac{t_2^2}{2} - 0} = \frac{2}{t_2}$ Using the orthogonality condition $m_{OP} \cdot m_{OQ} = -1$: $\left(\frac{2}{t_1}\right)\left(\frac{2}{t_2}\right) = -1 \implies t_1 t_2 = -4$ Thus, $t_2 = -\frac{4}{t_1}$.

2. Area of Triangle $\Delta OPQ$: The area of $\Delta OPQ$ with vertices $O(0,0)$, $P(x_1, y_1)$, and $Q(x_2, y_2)$ is given by: $\text{Area}(\Delta OPQ) = \frac{1}{2} |x_1 y_2 - x_2 y_1|$ Substituting the parametric coordinates of $P$ and $Q$: $\text{Area}(\Delta OPQ) = \frac{1}{2} \left| \frac{t_1^2}{2} t_2 - \frac{t_2^2}{2} t_1 \right| = \frac{1}{4} |t_1 t_2 (t_1 - t_2)|$ Since $t_1 t_2 = -4$ and $t_1 > 0 \implies t_2 < 0$, we have: $\text{Area}(\Delta OPQ) = \frac{1}{4} |-4| \left(t_1 - t_2\right) = t_1 - \left(-\frac{4}{t_1}\right) = t_1 + \frac{4}{t_1}$

3. Solving for $t_1$: Given that $\text{Area}(\Delta OPQ) = 3\sqrt{2}$: $t_1 + \frac{4}{t_1} = 3\sqrt{2}$ $t_1^2 - 3\sqrt{2}t_1 + 4 = 0$

Using the quadratic formula to find $t_1$: $t_1 = \frac{3\sqrt{2} \pm \sqrt{(3\sqrt{2})^2 - 4(1)(4)}}{2} = \frac{3\sqrt{2} \pm \sqrt{18 - 16}}{2} = \frac{3\sqrt{2} \pm \sqrt{2}}{2}$

This gives two possible values for $t_1$:

• Case 1: $t_1 = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$ $x = \frac{t_1^2}{2} = \frac{(2\sqrt{2})^2}{2} = 4, \quad y = t_1 = 2\sqrt{2}$ Hence, $P = (4, 2\sqrt{2})$.

• Case 2: $t_1 = \frac{2\sqrt{2}}{2} = \sqrt{2}$ $x = \frac{t_1^2}{2} = \frac{(\sqrt{2})^2}{2} = 1, \quad y = t_1 = \sqrt{2}$ Hence, $P = (1, \sqrt{2})$.

Therefore, the possible coordinates of $P$ are $(4, 2\sqrt{2})$ and $(1, \sqrt{2})$.

Correct Options: A and D