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The general equation for an ellipse is $Ax^2+Bxy+Cy^2+D=0$. How do I find the angle of rotation, the dimensions, and the coordinates of the center of the ellipse from the general equation and vice versa? Please avoid using matrices or parametric equations. I'd like all-in-one equations for each parameter.

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The general equation of an ellipse is: $$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$ if: $$4AC - B^2 > 0$$ The trick is to eliminate B so that the xy term vanishes.
If $B<>0$ then the ellipse is rotated and the angle of rotation is obtained from: $$tan(2 \theta) = \frac {B}{A-C}$$ $0 < \theta < \frac {\pi}{4}$

$$\cos {\theta} = \sqrt{\frac{1 + \cos{2 \theta}}{2}}$$ $$\sin{\theta} = \sqrt{\frac{1 - cos{2 \theta}}{2}}$$ Determine the new equation of the ellipse by calculating the following coefficients:
$A' = A \cos^2{\theta} + B \cos{\theta} \sin{\theta} + C \sin^2{\theta}$
$B' = 0$
$C' = A \sin^2{\theta} - B \cos{\theta} \sin{\theta} + C \cos^2{\theta}$
$D' = D \cos{\theta} + E \sin{\theta}$
$E' = -D \sin{\theta} + E \cos{\theta}$
$F' = F$

The resulting equation: $A'x'^2 + C'y'^2 + D'x' + E'y' + F' = 0$
After writing this equation in the form: $$ \frac{(x'-x'_0)^2}{a^2} + \frac{(y'-y'_0)^2}{b^2} = 1$$

We get: $$x'_0 = \frac{-D'}{2A'}$$ $$y'_0 = \frac{-E'}{2C'}$$

$$a^2 = \frac{-4F'A'C'+C'D'^2+A'E'^2}{4A'C'^2}$$ $$b^2 = \frac{-4F'A'C'+C'D'^2+A'E'^2}{4A'^2C'}$$

The coordinates of the centerpoint are found by rotating back about angle $\theta$

$x_0 = x'_0 \cos{\theta} - y'_0 \sin{\theta}$
$y_0 = x'_0 \sin{\theta} + y'_0 \cos{\theta}$


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The thing that is getting me with tan(2 * theta) = B / (A - C) is that I always get a result between 0 and pi/4 as you say. I was expecting a result between 0 and pi/2, because I want to know which of the axes (the major one or the minor one) is at the resulting angle, so that I can later draw it correctly. – Doug McClean Jul 2 '14 at 17:51
Maybe (eq. 23) gives an answer to that: if $a>c$, add $\pi/2$. – Torsten Bronger Jul 11 '15 at 13:13

$Ax^2+Bxy+Cy^2+Dx + Ey +F=0 \hspace{2 mm} .. (1)$ represents a general equation for conic. It includes a pair of straight line, circles, ellipse, parabola, and hyperbola. For this general equation to be an ellipse, we have certain criteria.

Suppose this is an ellipse centered at some point $(x_0, y_0)$. Our usual ellipse centered at this point is $$\frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} = 1 \hspace{ 2 cm } (2)$$

Note that this term does not have $xy$ term. This term appears due to rotation. So let's us counter rotate it such that $xy$ term vanishes. We let $x = X \cos \theta + Y \sin \theta$ and $y = - X \sin \theta + Y \cos \theta $, we plug this into $(1)$ and equate the coefficient of $xy$ to $0$ since we assume that by rotation $(xy)$ vanishes. From here, we can calculate $\theta $. If it's an ellipse then we are be able to reduce the remnant of $(1)$ into the form $(2)$. Hence we find the center.

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I learned how to rotate conics back in a course "calculus and analytic geometry" in the 1960s. I observe that more recent textbooks tend not to include it. It may (perhaps) be in linear algebra courses sometimes nowadays. – GEdgar Jan 17 '13 at 20:12
I do not really understand these transformation equations. GEdgar is right. I learned about ellipses in my trigonometry class and we only worked with ones aligned with axes. – Melab Jan 17 '13 at 23:24
it it explained here , sorry i had been writing it oppositely. fixed it – Santosh Linkha Jan 17 '13 at 23:33
What do $X$ and $Y$ equal? – Melab Jan 18 '13 at 0:07
They are the new coordinates after rotating in rotated coordinates. This red line becomes new X-axis and it's perpendicular new Y-axis – Santosh Linkha Jan 18 '13 at 11:28

The equation that you give is the equation of a general conic. Some of which are ellipses, some hyperbolae, some parabolae, and other degenerate conics, e.g. $xy=0$, $x^2=0$ or $x^2+y^2=0$.

The non-degenerate cases are distinguished by the way the conics interact with the line at infinity. We have:

  1. If $B^2-4AC < 0$ then we have an ellipse; the conic misses the line at infinity.
  2. If $B^2-4AC = 0$ then we have a parabola; the conic is tangent to the line at infinity.
  3. If $B^2-4AC > 0$ then we have a hyperbola; the conic crosses the line at infinity twice.

The centre of a conic is foung by solving the equations $\partial f/\partial x = \partial f /\partial y = 0$ where $f(x,y)$ is the equation you gave. In this case, assuming $B^2-4AC \neq 0$, we have the centre $(p,q)$ as:

$$(p,q) = \left(\frac{2CD-BE}{B^2-4AC},\frac{2AE-DB}{B^2-4AC}\right) . $$

As for the rotation, assume you have either an ellipse or a hyperbola. You need to complete the square on the $x^2$ and $x$ terms, as well as the $y^2$ and $y$ terms. This is equivalent to translating the conic so that its centre is at the origin. You can then re-label $(x-p)$ as $X$ and $(y-q)$ as $Y$ and divide through by a constant to give yourself a non-degenerate quadratic form, something like $aX^2+bXY + cY^2=1$ where $a$, $b$ and $c$ are real numbers. You then look at the matrix of the quadratic form:

$$Q = \left(\begin{array}{cc} a & \frac{1}{2}b \\ \frac{1}{2}b & c \end{array}\right) . $$

The eigenvectors of $Q$ give you the axes of the conic. In order to rotate the matrix, you need to find an orthogonal change of basis matrix which diagonalises $Q$. Finally, you will end up with $\alpha X^2+\beta Y^2=1$.

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I do not know what eigenvectors are and I don't see how the matrix correlates to the description of the ellipse. – Melab Jan 17 '13 at 23:20
@Melab If you type "eigenvector" into Google then you will find many pages. An eigenvector of a matrix is a vector whose direction is fixed by that matrix, i.e. $Mv = \lambda v$ where $M$ is a matrix, $v$ is a vector and $\lambda$ is a number. As for the matrix, multiply the left by $(X,Y)$ and the right by $(X,Y)^{\top}$. What do you get? You get $aX^2+bXY+cY^2$. – Fly by Night Jan 18 '13 at 18:57

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