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It is known that a conic can be expressed as an implicit second-order polynomial as follows: $$au^2 + buv + cv^2 + du + ev + f = 0$$ where $(u,v)$ is the 2d coordinate of the point on the conic. If we try to make it an ellipse, there is a discriminant $b^2-4ac<0$. My question is that how is this discriminant $b^2-4ac<0$ come from? Can anyone provide the derivation or give me some reference can reveal the answer? Thanks.

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Thank you for editing Thijs – ChengLu Aug 30 '11 at 15:44
up vote 2 down vote accepted

Since I still haven't seen the answer I wanted to see, I'll attempt writing that answer I was hoping somebody more eloquent than me would write, starting from my comment in Thijs's answer.

It's not too much trouble to determine the nature of a conic that doesn't have the "cross term" (the term with $uv$ in the OP's notation):


Easily, we see that if either of $a$ or $c$ are zero (but not both!), we get a parabola. If neither are zero, we can perform a further translation such that the conic is of the form


Geometrically speaking: if neither $a$ nor $c$ are zero, you have a "central conic" (an ellipse or hyperbola), and that "further translation" corresponds to changing coordinates such that the conic is now centered at the origin. In this form, checking whether you have an ellipse ($ac > 0$) or hyperbola ($ac < 0$) is easy as pie.

Since it's so, so convenient to check the nature of a conic expressed in these forms, one would want to be able to do a similar check for the case when the cross term has a nonzero coefficient. The trick is to rotate your coordinate system.

More precisely, consider the substitutions


Geometrically, this transformation rotates whatever it is applied to about the origin by the angle $\varphi$, anticlockwise. If we apply this to the form of the conic with the cross-term we obtain (ignoring non-quadratic terms for the remainder of this answer):

$$\begin{align*}&(a\cos^2\varphi+b\sin\,\varphi \cos\,\varphi+c\sin^2\varphi){u^\prime}^2+((c-a)\sin\,2\varphi+b\cos\,2\varphi)u^\prime v^\prime+\\&(a\sin ^2\varphi-b\sin\,\varphi \cos\,\varphi+c\cos^2\varphi){v^\prime}^2+\dots=0\end{align*}$$

One can then pick a rotation that zeroes out the cross term; this yields the equation


I'll omit the algebra and just say that you can derive an expression for $\tan\,\varphi$ from that:


from which it is easy to derive expressions for $\cos\,\varphi$ and $\sin\,\varphi$. Replacing $\cos\,\varphi$ and $\sin\,\varphi$ in the rotated conic with the appropriate expressions yields


The product of the two quadratic terms is seen to be equal to $ac-\frac14 b^2$; remembering that the quadratic is an ellipse if this product is positive, we have the inequality

$$4ac-b^2 > 0$$

and the traditional format of the inequality (as mentioned in the OP) is but the reverse of this.

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Is it possible that your expression for $\tan \varphi$ should be $\tan \Phi = \frac{c-a \pm \sqrt{(a - c)^2+ b^2}}{2 b}$? Or am I missing something? I used the double angle formula for the tangent function. – Dror Aug 15 '12 at 7:50
Nope, what I wrote is correct. What did you do? – J. M. Aug 15 '12 at 7:53
I solved the equation $\frac{b}{a-c} = \frac{2\tan \phi}{1-\tan^2 \phi}$. I double checked and it still seems to be different then your result - In my previous comment the denominator should be $b$. – Dror Aug 15 '12 at 8:03
Hmm, did you use the reversed form? – J. M. Aug 15 '12 at 8:06
Actually I used mathematica, now now verified my calculation by hand. – Dror Aug 15 '12 at 8:14

Your quadratic bivariate polynomial can be rewritten as $$\frac{-4 a c f+a e^2+b^2 f-b d e+c d^2}{b^2-4a c}+a \tilde{u}^2 + \frac{4 a c^2 }{4 a c - b^2} \tilde{v}^2$$ where $$\tilde{u} = u + v \frac{b}{2 a} + \frac{d}{2 a}$$ and $$\tilde{v} = \frac{-b d + 2 a e}{4 a c} + \frac{4 a c - b^2}{4 a c} v$$

From this change of variables, you see that if $4 a c - b^2 > 0 $ the conic defines an ellipse. The case $4 a c - b^2 < 0$ corresponds to a hyperbola. The limit $4 a c - b^2 \to 0$ requires a different change of variables.

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I think there is an easier solution. Basically, just compute the determinant of the Hessian matrix, and you'll get the discriminant.

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Welcome to Math.SE! Consider making your answer more detailed. How to Answer – Live Forever Jul 9 '14 at 21:08

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