AMC12 2015 A

AMC12 2015 A · Q12

AMC12 2015 A · Q12. It mainly tests Quadratic equations, Coordinate geometry.

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 $a + b$?

抛物线 $y = ax^2 - 2$ 和 $y = 4 - bx^2$ 与坐标轴相交于恰好四个点，这四个点是一个面积为 12 的风筝的顶点。$a + b$ 等于多少？

(A) 1 1

(B) 1.5 1.5

(D) 2.5 2.5

(E) 3 3

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): The $y$-intercepts of the two parabolas are $-2$ and $4$, respectively, and in order to intersect the $x$-axis, the first must open upward and the second downward. Because the area of the kite is $12$, the $x$-intercepts of both parabolas must be $-2$ and $2$. Thus $4a-2=0$ so $a=\frac{1}{2}$, and $4-4b=0$ so $b=1$. Therefore $a+b=1.5$.

答案（B）：两条抛物线的 $y$ 轴截距分别为 $-2$ 和 $4$。为了与 $x$ 轴相交，第一条必须开口向上，第二条必须开口向下。由于风筝的面积为 $12$，两条抛物线的 $x$ 轴截距都必须是 $-2$ 和 $2$。因此 $4a-2=0$，所以 $a=\frac{1}{2}$；并且 $4-4b=0$，所以 $b=1$。因此 $a+b=1.5$。

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