AMC12 2018 A

AMC12 2018 A · Q16

AMC12 2018 A · Q16. It mainly tests Quadratic equations, Circle theorems.

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?

以下哪个描述了曲线 $x^2 + y^2 = a^2$ 和 $y = x^2 - a$ 在实数 $xy$ 平面上相交恰好 3 个点的 $a$ 的取值集合？

(A) $a = \frac{1}{4}$ $a = \frac{1}{4}$

(B) $\frac{1}{4} < a < \frac{1}{2}$ $\frac{1}{4} < a < \frac{1}{2}$

(D) $a = \frac{1}{2}$ $a = \frac{1}{2}$

(E) $a > \frac{1}{2}$ $a > \frac{1}{2}$

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Solving the second equation for $x^2$ gives $x^2=y+a$, and substituting into the first equation gives $y^2+y+(a-a^2)=0$. The polynomial in $y$ can be factored as $(y+(1-a))(y+a)$, so the solutions are $y=a-1$ and $y=-a$. (Alternatively, the solutions can be obtained using the quadratic formula.) The corresponding equations for $x$ are $x^2=2a-1$ and $x^2=0$. The second equation always has the solution $x=0$, corresponding to the point of tangency at the vertex of the parabola $y=x^2-a$. The first equation has 2 solutions if and only if $a>\frac12$, corresponding to the 2 symmetric intersection points of the parabola with the circle. Thus the two curves intersect at 3 points if and only if $a>\frac12$.

答案（E）：由第二个方程解出 $x^2$ 得 $x^2=y+a$，代入第一个方程得到 $y^2+y+(a-a^2)=0$。关于 $y$ 的多项式可因式分解为 $(y+(1-a))(y+a)$，因此解为 $y=a-1$ 和 $y=-a$。（或者也可以用求根公式得到这些解。）对应的 $x$ 方程为 $x^2=2a-1$ 和 $x^2=0$。第二个方程总有解 $x=0$，对应于抛物线 $y=x^2-a$ 在顶点处的切点。第一个方程当且仅当 $a>\frac12$ 时有 2 个解，对应于抛物线与圆的 2 个对称交点。因此，两条曲线当且仅当 $a>\frac12$ 时有 3 个交点。

Topics