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The normal at any point P on the parabola y² = 8x makes an angle theta lying in (0, pi/4] with the positive x-axis. If the maximum possible distance of P from the origin is 2*sqrt(lambda), find lambda.

1. 5
2. 4
3. 8
4. 10

Correct answer: 5

Solution

For P = (2t², 4t), the normal slope is -t, so |t| <= tan(pi/4) = 1; OP is largest at |t| = 1, giving OP = 2*sqrt(5), hence lambda = 5.

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