# Quadratic equation of an ellipse and ellipse description

I found on Mathworld ( http://mathworld.wolfram.com/Ellipse.html ) that the quadratic equation:

$$ax^2 + 2bxy + cy^2 + 2dx + 2fy + g = 0$$

represent an ellipse only when, after defining:

$$\Delta = \left|\begin{array}{ccc} a&b&d\\b&c&f\\d&f&g\end{array}\right|$$ $$J = ac-b^2$$ $$I = a+c$$

the following conditions are met: $\Delta \neq 0$, $J > 0$ and $\Delta/I < 0$.

What I don't understand are the constraint other than $J > 0$. I've never seen these constraints before, and I really wonder why they are here.

Also, I noticed that the formulas to find the semi-axes and angle given after on the MathWorld article work only if these constraints are met.

The second part of my question is then: is there an other way to find the length and orientation of the axis of an ellipse given its quadratic equation? Or is it always possible to transform the quadratic equation of an ellipse into an equivalent one respecting the constraints described on MathWorld?

-
My understanding is that $J>0$ guarantees that the graph of the equation is an ellipse or a degenerate case of an ellipse, so I'd guess that the other conditions guarantee that it is not degenerate. – Isaac Jan 20 '11 at 18:32

The condition $\Delta/I < 0$ is necessary for the solution set to exist at all. For example, consider $x^2 + y^2 + 1 = 0$, which has no solutions; here $\Delta = 1$ and $I = 2$. I think the necessity of this condition can be shown using linear algebra; leave a comment if you'd like to see the proof.
Unless I'm missing something, $\Delta \neq 0$ is redundant, as it is implied by $\Delta/I < 0$.
My problem is, I have equations for which $\Delta/I$ > 0 but do correspond to an ellipse if plotted. I have the (bad?) luck to find it using ellipse fitting on a real set of data points. Or at least, the set exist and really look like an ellipse. – PierreBdR Jan 21 '11 at 11:18