Question

下列哪一项描述了实 $xy$ -平面中曲线 $x^2 + y^2 = a^2$ 和 $y = x^2 - a$ 恰好有 $3$ 个交点时 $a$ 的取值集合？

Which of the following describes the set of values of $a$ for which the curves $x^2 + y^2 = a^2$ and $y = x^2 - a$ in the real $xy$ -plane intersect at exactly $3$ points?

$a = \tfrac14$

$\tfrac14 \lt a \lt \tfrac12$

$a \gt \tfrac14$

$a = \tfrac12$

$a \gt \tfrac12$

答案：E

知识点：圆抛物线二次方程

难度评级：1840

解答：

将 $x^2 = y + a$ 代入 $x^2 + y^2 = a^2$ ，得 $y^2 + y + (a - a^2) = 0$ ，可分解为 $(y + 1 - a)(y + a) = 0$ ，所以 $y = a - 1$ 或 $y = -a$ 。它们分别对应 $x^2 = 2a - 1$ 和 $x^2 = 0$ 。

方程 $x^2 = 0$ 总是给出单个点 $(0, -a)$ ，即抛物线的顶点。方程 $x^2 = 2a - 1$ 恰好在 $2a - 1 \gt 0$ ，即 $a \gt \tfrac12$ 时给出另外两个点。因此恰好有 $3$ 个交点当且仅当 $a \gt \tfrac12$ 。

所以正确答案是 E。

Substituting $x^2 = y + a$ into $x^2 + y^2 = a^2$ gives $y^2 + y + (a - a^2) = 0,$ which factors as $(y + 1 - a)(y + a) = 0,$ so $y = a - 1$ or $y = -a.$ These correspond to $x^2 = 2a - 1$ and $x^2 = 0.$

The equation $x^2 = 0$ always gives the single point $(0, -a),$ the vertex of the parabola. The equation $x^2 = 2a - 1$ gives two more points exactly when $2a - 1 \gt 0,$ i.e. $a \gt \tfrac12.$ So there are $3$ intersection points precisely when $a \gt \tfrac12.$

Thus, the correct answer is E.

2018 AMC 12A 第 16 题

答案：E

难度评级：1840

解答：

其他年份的第 16 题