$B^2 - 4AC$ is called the discriminant of a conic section. It is an invariant. Depending on the sign of $B^2 - 4AC$, you can tell which of the three conic sections (Ellipse, Hyperbola, Parabola) where $A$, $B$, and $C$ are the coefficients of a rotated Conic Section is described by the equation

$$AX^2 + BXY + CY^2 + DX + EY + F = 0$$

Can anyone explain the logical reasoning to why $B^2 - 4AC$ can identify which Conic Section the equation is?

If $B^2 - 4AC$:

Is greater than zero = Hyperbola

Is less than zero = Ellipse

Is equal to zero = Parabola

  • 2
    $\begingroup$ Welcome. Is there any reason why we should obey your orders? ;-) Seriously, the question lacks context. What do you know about the matter? What are your thoughts and attempts? $\endgroup$ – egreg Jun 19 '16 at 23:43
  • 1
    $\begingroup$ What "derivation" are you looking for? Are you asking about a proof? A motivation? The expression $b^2 - 4ac$ should be familiar to you from high school algebra, so it would expedite responses if you explain what context your Question came from. $\endgroup$ – hardmath Jun 20 '16 at 2:20

There are two ways to prove this:

Formal: You can show, through a bunch of ugly computation, that the expression $B^2-4AC$ is invariant under rotation. So, consider when $B=0$ (in other words, when the conic section's directrix is parallel to one of the axes). It is easy to see that for a hyperbola $-4AC$ is positive, for an ellipse $-4AC$ is negative, and for a parabola $-4AC$ is $0$. For a better worded explanation, go to this link.

Very Informal But Intuitive: Take the equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$. Imagine if $x$ and $y$ were very large numbers. We can forget about $Dx+Ey+F$ because it becomes insignificant when compared to $Ax^2+Bxy+Cy^2=0$. Now, divide by $x^2$:


We notice that we now have a quadratic in $\frac{y}{x}$.

This next part is a little hard to explain in words (and my English is sort of bad) but I will try my best.

The number of solutions to this equation represents the number of ways in which the graph of the equation "zooms off towards infinity." Imagine zooming out really far from a graph of a hyperbola. You would only see an "X" formed by two lines (these lines are the asymptotes of the hyperbola). If you solve for $\frac{y}{x}$ in the above equation, you would be solving for the slopes of those lines. Imagine zooming out really far from a graph of a parabola. You would only see one line (the axis of symmetry for the parabola). If you solve for $\frac{y}{x}$ in the above equation, you would be solving for the slope of that line. If you zoomed out really far from a graph of an ellipse, you would see a point.

So, if $A+B\left(\frac{y}{x}\right)+C\left(\frac{y}{x}\right)^2=0$ has two solutions for $\frac{y}{x}$, the equation is a hyperbola. One solution means parabola. Zero solutions means ellipse or circle. The number of solutions corresponds to the sign of $B^2-4AC$.

I sort of like this informal proof because it explains why the discriminant of a conic looks like that of a quadratic.

  • 1
    $\begingroup$ I understand my second proof is very informal. If someone could edit and formalize my second proof, I would very much appreciate it. $\endgroup$ – Hrhm Jun 19 '16 at 23:53
  • $\begingroup$ Just wondering, is x/y a typo? Was it supposed to be y/x? $\endgroup$ – Alan Tam Jun 22 '16 at 18:05
  • $\begingroup$ @AlanTam Yeah, I'll fix that. $\endgroup$ – Hrhm Jun 22 '16 at 18:05
  • $\begingroup$ Also, when solving for y/x in the parabola, why would it be the slope of the axis? The parabola goes the infinity and the axis is not an asymptote. $\endgroup$ – Alan Tam Jun 22 '16 at 18:07
  • $\begingroup$ @AlanTam I believe it's because parabolas only have one point at infinity which is along the axis of symmetry (I need to confirm this). But a more intuitive way of seeing this is by plotting $y=x^2$ on a graphing calculator and zooming out really far. You'll only see one line, the axis of symmetry. $\endgroup$ – Hrhm Jun 22 '16 at 18:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.