2004 AMC 12B Problems/Problem 18

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 $AB$ ?

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

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 $\sqrt{ 1^2 + 7^2 } = \sqrt{50} = \boxed{5\sqrt2}$ .

2004 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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2004 AMC 12B Problems/Problem 18

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