Let P and Q be distinct points on the parabola y² = 2x such that a circle with PQ as diameter passes through the vertex O of the parabola. If P lies in the first quadrant and the area of the triangle ΔOPQ is 3√2, then which of the following is (are) the coordinates of P?

Parabola: y² = 2x
A point at parabola be P(t₁²/2, t₁), Q(t₂²/2, t₂)
A circle is drawn with PQ as diameter and passes through O(0, 0).
∠POQ = 90° in ΔOPQ
∴ PO and OQ are perpendicular.
∴ (Slope of OP) × (Slope of OQ) = -1
⇒ (2t₁/t₁²) × (2t₂/t₂²) = -1
⇒ 4/(t₁t₂) = -1
This is incorrect. The condition that the circle with PQ as diameter passes through O means that ∠POQ = 90°. The slope of OP is 2t₁/t₁² = 2/t₁, and the slope of OQ is 2/t₂. Since the lines are perpendicular, (2/t₁)(2/t₂) = -1, which simplifies to t₁t₂ = -4.
Let P = (t²/2, t) and Q = (s²/2, s). Then the area of triangle OPQ is given by
Area(OPQ) = 0.5 * |(t²/2)(s) - (s²/2)(t)| = 0.5 * |(ts/2)(t - s)| = 3√2
|ts(t - s)| = 12√2
Since t₁t₂ = -4, let t₁ = t and t₂ = s. Then ts = -4. Substituting this into the area equation gives:
|-4(t - s)| = 12√2
|t - s| = 3√2
We have the system of equations:

1. ts = -4
2. |t - s| = 3√2
   From (2), (t-s)² = 18, so t² - 2ts + s² = 18. Since ts = -4, we have t² + 8 + s² = 18, so t² + s² = 10.
   We also have (t + s)² = t² + 2ts + s² = t² + s² + 2ts = 10 + 2(-4) = 2, so t + s = ±√2.
   Then we can solve the quadratic t² ± √2t - 4 = 0 and s² ± √2s - 4 = 0.
   Solving these equations gives us the points (1, √2), (9, 3√2), (4, 2√2) and so on. However, since ts = -4, these points are P and Q.
   If P = (1, √2), then Q = (-4, -2√2), which isn't in the first quadrant, so P = (9, 3√2) or (4, 2√2). But only (9, 3√2) is an option.