2018 AMC 12A 第 16 题

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

小提示：

由

y = x^2 - a

将

x^2 = y + a

代入圆的方程，得到关于

y

的二次方程

From

y = x^2 - a

substitute

x^2 = y + a

into the circle equation to get a quadratic in

y

大提示：

一个交点是

x = 0

处的顶点相切；还要两个交点，则需要

x^2 = 2a - 1

有非零实数解。

One intersection is the vertex tangency at

x = 0;

two more require

x^2 = 2a - 1

to have real nonzero solutions

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

小提示：

大提示：

解答：

其他年份的第 16 题