Intersection of a circle $x^2+(y-1)^2=1$ and a parabola $ax^2=y$, bounds for a I have a circle $x^2+(y-1)^2=1$ and a parabola $x^2=\frac{y}{a}$ and it is required to find the bounds for $a$ if they intersect at point other than origin.
So I proceeded as follow, substituted $x^2=y/a$ into $x^2+(y-1)^2=1$
which led me to get,
$$\implies y^2+y\bigg(\frac{1}{a}-2\bigg)= 0$$
So for more than one point of intersection, I would necessarily have $D>0$, $D$ being the discriminant of the quadratic in $y$
From there I got the bounds as $a \in \bigg(R-\{0, \frac{1}{2}\}\bigg)$
But if I substitute the value of $y$ other than $0$, ie, $y = 2-\frac{1}{a}$ into $x^2=\frac{y}{a}$, I get $x^2 = \frac{2a-1}{a^2}$, which is always greater than $0$, and as a result from here I get the bound that $a \in \bigg(\frac{1}{2}, \infty\bigg)$

We can compute very easily the intersections of the two conics, or:
$$\frac{y}{a}+(y-1)^2=1 \leftrightarrow ay^2+y(1-2a)=0 \leftrightarrow y(ay+1-2a)=0$$
Now, a solution is $y=0$ that holds for evry $a$. The other solution is $y=\frac{2a-1}{a}$ and so $x=\frac{1}{a}\sqrt{2a-1}$. Now, taking from here, it's very simple. In fact we want the square root to be defined, so:
$$2a-1>0 \leftrightarrow a>\frac{1}{2}$$
Note that your error has been committed when stating tehe range of $a$ without considering the fact that $x=\pm\sqrt{\frac{y}{a}}$. In fact, as you stated, if $a\in\left(R-\left\{0,\frac{1}{2}\right)\right)$, then for example $a=-1$ would be OK, but if we plug into the equation for $x$ we have a contraddiction. In fact:
$$y(-y+3)=0 \leftrightarrow y=0\vee y=3$$
But now:
$$x=\pm\sqrt{\frac{y}{a}}=\pm\sqrt{\frac{3}{-1}}=\text{IMPOSSIBLE}$$
