AMC12 2004 B

AMC12 2004 B · Q18

AMC12 2004 B · Q18. It mainly tests Graphs (coordinate plane), Coordinate geometry.

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 $AB$?

点 $A$ 和 $B$ 在抛物线 $y=4x^2+7x-1$ 上，且原点是 $AB$ 的中点。$AB$ 的长度是多少？

(A) 2\sqrt{5} 2\sqrt{5}

(B) 5 + \frac{\sqrt{2}}{2} 5 + \frac{\sqrt{2}}{2}

(D) 7 7

(E) 5\sqrt{2} 5\sqrt{2}

Answer

Correct choice: (E)

正确答案：（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 $\sqrt{ 1^2 + 7^2 } = \sqrt{50} = \boxed{5\sqrt2}$.

设 $A$ 的坐标为 $(x_A,y_A)$。由于 $A$ 在抛物线上，有 $y_A=4x_A^2+7x_A-1$。由于原点是 $AB$ 的中点，点 $B$ 的坐标为 $(-x_A,-y_A)$。我们需要选择 $x_A$ 使得 $B$ 也在抛物线上。换句话说，需要满足 $-y_A = 4(-x_A)^2 + 7(-x_A) - 1$。代入 $y_A$，得到：$-4x_A^2 - 7x_A + 1 = 4(-x_A)^2 + 7(-x_A) - 1$。化简得 $8x_A^2 - 2 = 0$，解得 $x_A = \pm 1/2$。两个解对应同一对点：$(1/2,7/2)$ 和 $(-1/2,-7/2)$。它们的距离为 $\sqrt{ 1^2 + 7^2 } = \sqrt{50} = \boxed{5\sqrt2}$。

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