Question

抛物线 $y = ax^2 - 2$ 和 $y = 4 - bx^2$ 与坐标轴恰好交于四个点，这四个点是一只面积为 $12$ 的风筝形的顶点。求 $a + b$ ？

The parabolas $y = ax^2 - 2$ and $y = 4 - bx^2$ intersect the coordinate axes in exactly four points, and these four points are the vertices of a kite of area $12.$ What is $a + b?$

$1$

$1.5$

$2$

$2.5$

$3$

答案：B

知识点：抛物线筝形坐标几何面积

难度评级：1630

解答：

两条抛物线的 $y$ -截距为 $-2$ 和 $4$ 。为了与 $x$ -轴相交，第一条抛物线开口向上，第二条开口向下，所以它们的 $x$ -截距为 $\pm t$ ，其中 $t \gt 0$ 。

这个风筝形沿 $y$ -轴的一条对角线长度为 $4 - (-2) = 6$ ，另一条长度为 $2t$ 。面积为 $\dfrac{1}{2}\cdot 6\cdot 2t = 6t = 12$ ，所以 $t = 2$ 。

因此 $x$ -截距为 $\pm 2$ 。对第一条抛物线， $0 = a(2)^2 - 2$ 得 $a = \dfrac{1}{2}$ ；对第二条， $0 = 4 - b(2)^2$ 得 $b = 1$ 。因此 $a + b = 1.5$ 。

因此，正确答案是 B。

The $y$ -intercepts of the two parabolas are $-2$ and $4.$ To intersect the $x$ -axis, the first parabola opens upward and the second opens downward, so their $x$ -intercepts are $\pm t$ for some $t \gt 0.$

The kite has one diagonal of length $4 - (-2) = 6$ along the $y$ -axis and the other of length $2t.$ Its area is $\dfrac{1}{2}\cdot 6\cdot 2t = 6t = 12,$ so $t = 2.$

Thus the $x$ -intercepts are $\pm 2.$ For the first parabola, $0 = a(2)^2 - 2$ gives $a = \dfrac{1}{2};$ for the second, $0 = 4 - b(2)^2$ gives $b = 1.$ Therefore $a + b = 1.5.$

Thus, the correct answer is B.

2015 AMC 12A 第 12 题

答案：B

难度评级：1630

解答：

其他年份的第 12 题