Consider the parabola $y=x^2$ and an ellipse which ‘rests’ on it, given by the equation $$\frac{x^2}{a^2} +\frac{(y-h)^2}{b^2}=1$$The goal is to find all ordered pairs $(a,b)$ for which the ellipse doesn’t fall to the origin, namely it touches the parabola at two distinct points.
Replacing $y$ by $x^2$ in the equation of the ellipse, we get $$\frac{x^2}{a^2} +\frac{(x^2-h)^2}{b^2}=1 $$
I could calculate the discriminant of this quadratic in $x^2$ and set it $\gt 0$. Is there a quick and neat way to express $h$ in terms of $a,b$? Or maybe another approach to solve this problem?
I'd look at the problem in a different way. Suppose we are given $b = h > 0$ such that the ellipse touches the parabola's vertex. What is the largest $a$, say $a^*$, such that the ellipse has no other intersection points with the parabola? For $a > a^*$, the ellipse cannot have $h = b$, thus such an $(a,b)$ pair cannot "fall" all the way down to the vertex.
As such, we require $$\frac{x^2}{a^2} + \frac{(y-b)^2}{b^2} = 1, \\ y = x^2,$$ hence $$a^2 y^2 + (b^2 - 2a^2 b) y = 0,$$ for which the unique nonzero root in $y$ is $$y = \frac{(2a^2 - b)b}{a}.$$ Therefore, if this root exists and is positive, the ellipse has another intersection point with the parabola other than its vertex; i.e., if $2a^2 > b$ or $$a > a^* = \sqrt{b/2}.$$ It follows that the set of all ordered pairs for which the ellipse cannot touch the vertex are those pairs for which $a > \sqrt{b/2}$.
By request, there is the question of how to express $h$ as a function of $(a,b)$ for ellipses satisfying the condition $a > \sqrt{b/2}$. This involves only a slight modification of the above calculation, namely we solve the system $$\frac{x^2}{a^2} + \frac{(y-h)^2}{b^2} = 1, \\ y = x^2$$ for $y$, giving $$y = h - \frac{b^2}{2a^2} \pm \frac{b \sqrt{4a^4 + b^2 - 4a^2 h}}{2a^2}.$$ Then we note that under the assumptions $a > \sqrt{b/2}$ and $h \ge b$, we have $h - b^2/(2a^2) > h - b > 0$. When the ellipse is tangent to the parabola, the solution has a unique double root; i.e., the the discriminant $4a^2 + b^2 - 4a^2 h$ must be zero, or $$h = a^2 + \frac{b^2}{4a^2}.$$ This characterizes the location of such an ellipse when $h \ge b$.