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Let P be the point on the parabola y² = 4x that is nearest to the centre S of the circle x² + y² - 4x - 16y + 64 = 0. Let Q be the point on the circle that lies on segment SP. Which of the following is correct?

1. SP = 2*sqrt(5)
2. SQ: QP = (sqrt(5) + 1): 2
3. the x-intercept of the normal to the parabola at P is 6
4. the slope of the tangent to the circle at Q is 1/2

Correct answer: SP = 2*sqrt(5)

Solution

Minimising SP gives the nearest point P = (4, 4), so SP = sqrt(4 + 16) = 2*sqrt(5).

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