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In Keith Kendig's paper, Stalking the Wild Ellipse (published in the American Mathematical Monthly, November 1995), he says that if $A, B, C$ are chosen at random, the probability that the Cartesian equation $Ax^2+Bxy+Cy^2 = 1$ defines an ellipse is about $0.19$. How does one make this precise?

I assume that this statement is similar to, for example, the idea that the probability that "two random integers" are relatively prime is $\frac{6}{\pi^2}$. We choose uniformly from the range $1$ to $N$ and then look at the limit as $N\to\infty$.

So for the ellipse problem, do we choose $A,B,C$ from an $N \times N \times N$ cube (centered around the origin) and then compute a triple integral with $N\to\infty$?

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    Thanks Qiaochu, this is the nut of my question, which I perhaps didn't explain well.2012-05-02

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Here's what Kendig actually says:

Let's define "choosing $A,B,C$ at random" to mean picking an arbitrary point in a coordinate box in $(A,B,C)$-space, centered at the origin. A point corresponds to an ellipse exactly when $A\gt0$, $C\gt0$, and $B^2-4AC\lt0$. Show that the proportion of this box corresponding to an ellipse in the $(x,y)$-plane is ${31-3\log4\over144}=.1864$ Note that since $B^2-4AC$ is homogeneous, $B^2-4AC=0$ is a cone, and therefore this number is independent of the size of the box.

What part of this is not precise?

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    Would the down-voter please explain? Gerry has correctly answered the question that was asked, even if it now appears from the OP's comment that it wasn't the question that was intended.2012-05-02