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Find the locus of the trisection point of an arbitrary double ordinate of the parabola x² = 4by.

1. 9x² = by
2. 3x² = 2by
3. 9x² = 4by
4. 9x² = 2by

Correct answer: 9x² = 4by

Solution

Endpoints of a double ordinate are (h, k) and (-h, k) with h² = 4bk. A trisection point is (h/3, k); substituting h = 3x, k = y into h² = 4bk gives 9x² = 4by.

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