2004 AMC 12B Problem 18

Below is the professionally curated solution for Problem 18 of the 2004 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 12B solutions, or check the answer key.

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Concepts:parabola midpoint distance formula

Difficulty rating: 1740

18.

Points $A$ and $B$ are on the parabola $y = 4x^2 + 7x - 1,$ and the origin is the midpoint of $\overline{AB}.$ What is the length of $AB?$

$2\sqrt{5}$

$5 + \dfrac{\sqrt{2}}{2}$

$5 + \sqrt{2}$

$7$

$5\sqrt{2}$

Solution:

Let $B = (a, b)$ and $A = (-a, -b).$ Then $4a^2 + 7a - 1 = b$ and $4a^2 - 7a - 1 = -b.$ Subtracting gives $14a = 2b,$ so $b = 7a.$ Then $4a^2 + 7a - 1 = 7a$ gives $a^2 = \dfrac14,$ and $b^2 = 49a^2 = \dfrac{49}{4}.$ So $AB = 2\sqrt{a^2 + b^2} = 2\sqrt{\dfrac{50}{4}} = 5\sqrt{2}.$

Thus, the correct answer is E.

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