Interesting Parabolas

We note that $x^2+y^2 = a^2$ is a circle with center at the origin and a radius $a$ . While $y=x^2-a$ is a parabola with its vertex at $(0,-a)$ , which is on the circle (its lowest point). Depending on the value of $a$ , there are either 1 or 3 points where the circle and parabola meet. Let us check where the curves meets as follows.

$\begin{aligned} {\color{#3D99F6}x^2}+y^2 & = a^2 & \small \color{#3D99F6} \text{Given that }y=x^2-a \\ {\color{#3D99F6}y+a}+y^2 & = a^2 & \small \color{#3D99F6} \implies x^2 = y+a \\ y^2 + y + a+a^2 & = 0 & \small \color{#3D99F6} \text{Solving the quadratic} \end{aligned}$

$\implies y = \dfrac {-1 \pm \sqrt{1+4a^2-4a}}2 = \dfrac {-1\pm (2a-1)}2 = \begin{cases} a-1 \\ -a \end{cases}$

Note that $y_0=-a$ is the lowest point $(0,-a)$ , the curves meet. Let $y_1 = a-1$ . Then we have:

$\begin{aligned} y_1 & = a-1 \\ x_1^2 - a & = a-1 \\ x_1 & = \pm \sqrt{2a-1} \end{aligned}$

Note that $x_1$ is real only when $a \ge \frac 12$ . When $a < \frac 12$ , $y_1$ does not exist and the curves meet only at $(0,-a)$ . When $a=\frac 12$ , $a-1=a$ $\implies y_1=y_0$ also the curves meet only at $(0,-a)$ . When $\boxed{a>\frac 12}$ , the curves meet at $(0,-a)$ and $(\pm \sqrt{2a-1}, a-1)$ .

Solution:

Substituting $y=x^2-a$ into $x^2+y^2=a^2$ , we get $x^2+(x^2-a)^2=a^2 \implies x^2+x^4-2ax^2=0 \implies x^2(x^2-(2a-1))=0$ Since this is a quartic, there are 4 total roots (counting multiplicity). We see that $x=0$ always at least one intersection at $(0,-a)$ (and is in fact a double root).

The other two intersection points have $x$ coordinates $\sqrt{2a-1}$ . We must have $2a-1> 0$ , otherwise we are in the case where the parabola lies entirely above the circle (tangent to it at the point $(0,a)$ ). This only results in a single intersection point in the real coordinate plane. Thus, we see $\boxed{\textbf{(E) }a>\frac12}$ .