What causes the extraneous root when intersecting parabola $y^2=4ax$ with circle $x^2+y^2=9a^2/4$?

Question

If I solve the parabola: $y^2=4ax$ and the circle: $x^2+y^2=\frac{9a^2}{4}$ I get a quadratic in $x$; ie: $$4x^2+16ax-9a^2=0$$ which has roots $x=\frac{a}{2},-\frac{9a}{2}$. But if we see the graphs of these two, both the intersections occur at same $x$-coordinate so why does this extra root $x=-\frac{9a}{2}$ come?

So my question is that if the circle was unknown say $x^2+y^2=b^2$ then I would have put the discriminant of the quadratic in $x$ as $0$ but that would be wrong so how do I judge that an extraneous root will be present and discriminant won't be zero?

Answer 1

Basically when you combine the two equations, you have the implication that a solution of the two equations is also a solution of the combined equation but not necessarily the other way round! In this case you lose the information that $4ax\geq0$ which is implicitely given in the equation of the parabola.

edit: With regards to your second question. If you substitute $y^2$ you will always end up with 2 distinct roots for the $x$-equation regardless of the value of $b$ (unless $a=b=0$). It really depends on your choice of how you tackle the problem. E.g. if you substitute for $x$ (not that I would personally recommend it) you get a forth order equation: $$\frac{y^4}{(4a)^2}+y^2=\frac{9a^2}{4}$$ which yields the two real solutions for $y$ you want. That's why it is so important to doublecheck solutions after a computation and go back to the original problem.

Answer 2

You will need to reject $x = -\frac{9a}{2}$ since by the first equation, $a$ and $x$ must be of the same sign or zero in order to have $4ax \ge 0$.

Answer 3

Assuming that $a$ is a real number you would obtain: $$ y^2=4a\left(-\frac{9a}2\right)=-18a^2, $$ which has no real solutions for $y$, unless $a=0$, in which case both solutions degenerate to a single point $(0,0)$.

Answer 4

The full solution is this $$\begin{cases}x^2+y^2=\frac94a^2\\ y^2=4ax\end{cases}\iff \begin{cases}x^2+4ax-\frac94a^2=0\\ y^2=4ax\end{cases}\iff\begin{cases}x=\frac a2\\ y^2=2a^2\end{cases}\lor\begin{cases}x=-\frac 92a\\ y^2=-18a^2\end{cases}$$

Now the system must be "discussed" in terms of the parameter, in the sense that it is equivalent to the following: $$\begin{cases}x=\frac a2\\ y^2-2a^2=0\end{cases}\lor\begin{cases}x=-\frac 92a\\ y^2+18a^2=0\end{cases}\iff\begin{cases}x=\frac a2\\ y=\sqrt2a\end{cases}\lor\begin{cases}x=\frac a2\\ y=-\sqrt2a\end{cases}\lor\begin{cases}x=0\\ y=0\\ a= 0\end{cases}$$

And you decide whether or not to accept the degenerate case.

Answer 5

I don’t view $-9a/2$ as an “extra” root. If you allow $x$ and $y$ to take on complex values, then, counting repetitions, the system of second-degree equations that represents a pair of nondegenerate conics will have four solutions. Now, you’re looking for real points of intersection of those two conics, so any solutions that involve complex numbers are indeed extraneous in that context, but there are other applications in which knowing those “imaginary” intersection points is useful.

In your case, the solution $x=a/2$ gives rise to two real solutions $y=\pm\sqrt2a$ (which coincide in the degenerate $a=0$ case), while $x=-9a/2$ produces the imaginary solutions $y=\pm i3\sqrt2a$. Since you’re looking for real intersection points, you know to reject the latter value of $x$.