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A point P lies on the parabola y² = 4ax, where a > 0. The normal at P intersects the x-axis at a point Q. The triangle formed by P, Q, and the focus F of the parabola has an area of 120. If both the slope m of the normal and the value of a are positive integers, what is the pair (a, m)?

1. (2, 3)
2. (1, 3)
3. (2, 4)
4. (3, 4)

Correct answer: (2, 3)

Solution

With P=(a t^2, 2a t), the normal meets the x-axis at Q=(2a + a t^2, 0) and F=(a,0). The triangle area = (1/2)*a(1+t^2)*2a|t| = a^2 |t| (1+t^2). Setting m=|t| and a^2 m (1+m^2) = 120 gives the integer solution a=2, m=3 (option index 0); the stored (1,3) gives only 30.

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