# Ellipse equation parameters

If I have an ellipse expressed by: $ax^2 + 2cxy + by^2 = constant$

what does this expression equal to: $1/ \sqrt{ab - c^2}$ with respect to the ellipse ?

Thanks

matlabit

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is it an ellipse or an ellipsoid? $z$ can't be constant for an ellipsoid. – ajay Sep 12 '12 at 11:23
which equation?? – Aang Sep 12 '12 at 12:25
Where you write "equation", this is an expression, not an equation. Also note that you don't need the parentheses inside the square root. – joriki Sep 12 '12 at 13:22

[Note:] As David pointed out, I'm assuming the constant is $1$; if not, divide through by the constant.

This is the product of the two semi-axes, and thus except for a factor of $\pi$ it is the area of the ellipse.

Write the left-hand side as

$$\pmatrix{x&y}\pmatrix{a&c\\c&b}\pmatrix{x\\y}\;.$$

We can diagonalize the matrix to bring this into the form

$$\pmatrix{x'&y'}\pmatrix{\lambda_1&0\\0&\lambda_2}\pmatrix{x'\\y'}\;,$$

where $\lambda_1$ and $\lambda_2$ are the eigenvalues of the matrix and the inverse squares of the semi-axes. Their product is the determinant $ab-c^2$.

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You are assuming matlabit's "constant" is 1. – David Speyer Sep 12 '12 at 14:31

Why not divide the equation by ${\rm constant}$ to make it $a x^2+2 c x y + b y^2 = 1$ without any loss of generality.

Then for a rotated ellipse with major radius $r_1$ and minor radius $r_2$ and orientation angle $\theta$ the coefficients are

$$a = (\frac{1}{r_1^2}-\frac{1}{r_2^2}) \cos^2\theta+\frac{1}{r_2^2}$$ $$b = (\frac{1}{r_2^2}-\frac{1}{r_1^2}) \cos^2\theta+\frac{1}{r_1^2}$$ $$c = (\frac{1}{r_1^2}-\frac{1}{r_2^2}) \sin\theta \cos\theta$$

and the said quantity

$$\boxed{ \frac{1}{\sqrt{a b-c^2}} = r_1 r_2 }$$

which appears to be the area over $\pi$.

Note the equation of the ellipse is best expressed by the quadratic form of the conic section tensor $C$

$$\boldsymbol{x}^\top C \boldsymbol{x} = 0$$ $$\begin{pmatrix} x & y & 1 \end{pmatrix} \begin{pmatrix} a & c & 0\\c & b & 0\\0 & 0 & \text{-}1 \end{pmatrix} \begin{pmatrix} x\\y\\1 \end{pmatrix} = a x^2+2 c x y + b y^2 -1 = 0$$

or in rotated coordinates $x'=x \cos\theta-y\sin\theta$, $y'=x \sin\theta+x\cos\theta$

$$\begin{pmatrix} x' & y' & 1 \end{pmatrix} \begin{pmatrix} \frac{1}{r_1^2} & 0 & 0\\0 & \frac{1}{r_2^2} & 0\\0 & 0 & \text{-}1 \end{pmatrix} \begin{pmatrix} x'\\y'\\1 \end{pmatrix} = \frac{x'^2}{r_1^2}+\frac{y'^2}{r_2^2}-1 = 0$$

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Thank You guys, both answers were really helpful! – matlabit Sep 13 '12 at 8:49