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For the curve y² = 2x³, identify the point P at which the tangent is perpendicular to the line 4x - 3y + 2 = 0.

1. (2, 4)
2. (1, √2)
3. (1/2, 1/2)
4. (1/8, -1/16)

Correct answer: (1/8, -1/16)

Solution

Differentiate y^2=2x^3: 2y y' = 6x^2 -> y' = 3x^2/y. The line 4x-3y+2=0 has slope 4/3, so the perpendicular tangent has slope -3/4: 3x^2/y = -3/4 -> y = -4x^2. Substituting into y^2=2x^3 gives 16x^4 = 2x^3 -> x = 1/8, y = -1/16. Point P = (1/8, -1/16).

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