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Let P and Q be distinct points on the parabola y² = 2x such that the circle with PQ as diameter passes through the vertex O of the parabola. If P is in the first quadrant and the area of triangle OPQ is 3*sqrt(2), then the coordinates of P can be:

1. (4, 2*sqrt(2))
2. (9, 3*sqrt(2))
3. (1/4, 1/sqrt(2))
4. (1, sqrt(2))

Correct answer: (4, 2*sqrt(2))

Solution

With P = (p²/2, p) and Q = (q²/2, q), perpendicularity gives pq = -4 and the area gives |p - q| = 3*sqrt(2), yielding p = 2*sqrt(2) (so P = (4, 2*sqrt(2))) or p = sqrt(2) (so P = (1, sqrt(2))). Option (4, 2*sqrt(2)) is one valid choice.

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