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2004 AMC 12B Problems/Problem 18

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Problem

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

\mathrm{(A)}\ 2\sqrt5\qquad\mathrm{(B)}\ 5+\frac{\sqrt2}{2}\qquad\mathrm{(C)}\ 5+\sqrt2\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)...

Solution

Let the coordinates of A be (x_A,y_A). As A lies on the parabola, we have y_A=4x_A^2+7x_A-1. As the origin is the midpoint of AB, the coordinates of B are (-x_A,-y_A). We need to choose x_A so that B will lie on the parabola as well. In other words, we need -y_A = 4(-x_A)^2 + 7(-x_A) - 1.

Substituting for y_A, we get: -4x_A^2 - 7x_A + 1 = 4(-x_A)^2 + 7(-x_A) - 1.

This simplifies to 8x_A^2 - 2 = 0, which solves to x_A = \pm 1/2. Both roots lead to the same pair of points: (1/2,7/2) and (-1/2,-7/2). Their distance is \sqrt{ 1^2 + 7^2 } = \sqrt{50} = \boxed{5\sqrt2}.

See Also

2004 AMC 12B (ProblemsResources)
Preceded by
Problem 17
Followed by
Problem 19
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Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
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