Intersection of a circle $x^2+(y-1)^2=1$ and a parabola $ax^2=y$, bounds for a

Question

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 $a \in \bigg(\frac{1}{2}, \infty\bigg)$

Answer 1

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: $$x=\pm\sqrt{\frac{y}{a}}=\pm\sqrt{\frac{3}{-1}}=\text{IMPOSSIBLE}$$

Answer 2

What if $a=0$?

Any point on the circle is $\cos t,1+\sin t$

$$a(1-\sin^2t)=1+\sin t$$

$$\iff a\sin^2t+\sin t+1-a=0$$

$$\sin t=\dfrac{-1\pm\sqrt{1-4a(1-a)}}{2a}=\dfrac{-1\pm(2a-1)}{2a}=1-\dfrac1a, -1$$

So, for any finite value of $a, (0,0)$ is the first intersection

So for second intersection, $$-1\le1-\dfrac1a\le1\iff-2\le-\dfrac1a\le0$$

$$-\dfrac1a\le0\implies a>0$$

$$-2\le-\dfrac1a\iff2\ge\dfrac1a\implies a\ge\dfrac12\text{ as } a>0$$