Question

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?

$a = \tfrac14$

$\tfrac14 \lt a \lt \tfrac12$

$a \gt \tfrac14$

$a = \tfrac12$

$a \gt \tfrac12$

Solution:

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 Problem 16

Solution:

Problem 16 in Other Years