Given the circle $x^2+y^2=r^2$ and the parabola $y=ax^2+\frac{5}{3}r$, where $r>0$, find the coefficient $a$ such that the parabola is tangent to the circle.

Setting up a system with the two equations and letting the discriminant equal to zero yelds another second-degree equation, namely \begin{equation} 1+4a^2r^2+\frac{20}{3}ar = 0 \end{equation} which, when solved, gives two possible values for $a$: \begin{equation} a=-\frac{3}{2r}\qquad\text{and}\qquad a=-\frac{1}{6r} \end{equation} Now, only the first one is the actual solution. My question is, why does the second solution even show up? I was expecting just one possibility for $a$, given the geometry of the problem. What does the other solution mean? Does it have anything to do with complex numbers?


I found your same result, indeed from the condition we obtain

$$y^2+\frac y a - \frac 5 3 \frac r a -r^2=0$$

which leads to

$$\frac 1 {a^2}+\frac {20} 3\frac r a +4 r^2=0$$

and then

$$\frac 1 a=\frac{-\frac{20}3 r \pm \frac{16}3 r}{2} \implies a=-\frac{3}{2r}\qquad\text{and}\qquad a=-\frac{1}{6r}$$

The second solution is probably added when we take the condition $\Delta =B^2-4AC=0$.

  • $\begingroup$ ok but does it have any meaning? It should have, in my opinion... $\endgroup$ – marco trevi May 15 '18 at 17:56
  • $\begingroup$ See my answer. The equation you were solving had nothing to do with finding tangent curves, unfortunately. In a sense, you got lucky that it gave you a right answer because of other details of the geometry of the problem. $\endgroup$ – Steve Kass May 15 '18 at 18:26

Steve Kass is correct that having only one solution in $y$ doesn't itself imply tangency. But given the geometry of the problem, we can give that a pass. Instead of focusing on the tangency, let's redirect the problem:

Suppose we want the two curves

\begin{cases} x^2 + y^2 = r^2 \\ y^2 = ax^2 + \dfrac35 r \end{cases}

to intersect in exactly two points, what's the relationship between $a$ and $r$?

On to the actual answer

Does it have anything to do with complex numbers?

Kind of. Notice the equations are even in $x$, so we are actually solving it for $x^2$ and $y$. Substituting $u = x^2$ gives it a more convenient form

\begin{cases} u + y^2 = r^2 \\ y^2 = au + \dfrac35 r \end{cases}

And we want this system to have only one solution pair in $(u,y)$. When you actually plot these two curves in the new coordinate system, you'll see that the "real" solution $a = -\frac{3}{2r}$ gives an intersection point where $u > 0$, while the "fake" solution gives an intersection point where $u < 0$.

enter image description here

In short, both solutions work, but one is discarded when we restrict the domain to $u \ge 0$


The short answer, unfortunately, is that your approach to solving the problem is flawed and you got lucky...

It looks like you set the two equations equal to each other to solve for $y$. This will help you find possible $y$-coordinate values where the circle and parabola equations are both satisfied. If such a $y$-values is real, the curves will have at least one intersection point at that $y$-value. If such a $y$-value is not real, the equations are satisfied simultaneously for that $y$, but it has nothing to do with the graphs in the real plane. (In any case, finding $y$-values where both equations are satisfied has nothing to do with tangency.)

Setting the discriminant equal to zero will let you find values of $a$ for which there is a unique $y$-coordinate value satisfying both equations. This still has nothing to do with tangency. If you plot the two curves for $a=\frac{-3}{2r}$, you will see that the curves intersect at two points with the same $y$-value but opposite $x$-values.

enter image description here

You are a bit lucky to have found the $a$-value where the curves are tangent. If the point of tangency were at the parabola vertex and the parabola were “narrow” and curved in the same direction of the circle, there would have been three intersection points with two different $y$-coordinates, and you wouldn’t have found the solution.

enter image description here

  • $\begingroup$ but of course setting the discriminant to zero amounts to finding tangents...the sign of the $\Delta$ tells you if there are or not real solutions, i.e. intersection points...when different points coincide, i.e. when $\Delta=0$, we have tangency...or not? $\endgroup$ – marco trevi May 15 '18 at 19:00
  • $\begingroup$ This is a good analysis, but it still doesn't answer the question of why there is an extra solution. Ignoring the tangency for a second an focus on intersections only; the second solution does not produce any intersection. $\endgroup$ – Dylan May 15 '18 at 20:30
  • $\begingroup$ Setting a quadratic $f(z)$’s discriminant to zero finds values of a parameter for which $f(z)=0$ has one solution, not two. In your case, it is not true that “solutions” and “intersection points” are the same. Your quadratic’s roots are the $y$-coordinates of intersection points (if real - if they are complex, they are unrelated to the graph). Why is there an extra solution?” isn’t the right question to ask. The equation you’re solving is not the right equation for finding intersection points, so the solution you’re asking about isn’t “extra.” It’s a valid solution to a different question. $\endgroup$ – Steve Kass May 15 '18 at 20:58
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    $\begingroup$ But the first equation is the right equation for intersection points. The second equation came from the condition of wanting only one solution in $y$. The actual problem is one of the solutions (of the parameter $a$) produces imaginary $x$-coordinates. Hence, an "extra" solution that is no longer valid due to other constraints $\endgroup$ – Dylan May 15 '18 at 21:06
  • $\begingroup$ To explain why the “second solution doesn’t produce an intersection”: With the second solution, the two equations do have simultaneous solutions for a unique $y$-value. (That’s what setting the discriminant to zero gives.) However, for that $y$-value, $x$ is imaginary, so that $y$-value is not the $y$-coordinate of a point on the graphs. The graphs represent the solutions of the equations for real $x$ and $y$ only. Analogy: $y=x^2+1$ and $y=0$ have a common solution when $y=0$, but the graphs don’t intersect, because for $y=0$, $x$ is imaginary. Plug in the second solution and solve to see! $\endgroup$ – Steve Kass May 15 '18 at 21:17

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