# Finding the angle of rotation of an ellipse from its general equation and the other way around

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.

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

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

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

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

Cheers!

• 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. Jul 2, 2014 at 17:51
• Maybe mathworld.wolfram.com/Ellipse.html (eq. 23) gives an answer to that: if $a>c$, add $\pi/2$. Jul 11, 2015 at 13:13
• @JThemen What about vice versa of the question? May 11, 2016 at 7:43
• @TorstenBronger beware of wolfram - they write the general quadratic as $Ax^2 + 2Bxy + CY^2 + 2Dx + 2Ey + F$ Jan 22, 2020 at 17:29

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 $\alpha X^2+\beta Y^2=1$.

EXAMPLE

Take the example $10x^2+12xy+10y^2=1$. The quadratic form on the left has matrix $$M := \left(\begin{array}{cc} 10 & 6 \\ 6 & 10 \end{array}\right)$$ The eigenvectors of $M$ are $(1,1)^{\top}$ and $(1,-1)^{\top}$ with corresponding eigenvalues of $16$ and $4$ respectively. That gives $(1/\sqrt 2,1/\sqrt 2)^{\top}$ and $(1/\sqrt 2, -1/\sqrt 2)^{\top}$ as an orthogonal basis, i.e. unit length and perpendicular. Changing to this basis will diagonalise the quadratic form. (The change of basis will be a rotation.) If $$V:=\left(\begin{array}{cc} 1/\sqrt 2 & 1/\sqrt 2 \\ 1/\sqrt 2 & -1/\sqrt 2\end{array}\right) \ \ \ \mbox{and} \ \ \ D:=\left(\begin{array}{cc} 16 & 0 \\ 0 & 4 \end{array}\right)$$ then $MV=VD$. The matrix $V$ is orthogonal, and so $V^{-1}=V^{\top}$ meaning that $D=V^{-1}MV=V^{\top}MV$. If $x = {x\choose y}$ then \begin{eqnarray*} x^{\top}D x &=& x^{\top}V^{\top}MV x \\ \\ &=& (V x)^{\top}M(V x) \end{eqnarray*} A change of coordinates $x \mapsto Vx$ will diagonalise the quadratic form. If we put \begin{eqnarray*} x &=& \frac{1}{\sqrt 2}\,X + \frac{1}{\sqrt 2}\,Y \\ \\ y &=& \frac{1}{\sqrt 2}\, X - \frac{1}{\sqrt 2}\, Y \end{eqnarray*} then $10x^2+12xy+10y^2 \leadsto 16X^2 + 4Y^2$. Meaning that the locus $10x^2+12xy+10y^2=1$ can be rotated to the locus $16x^2+4y^2=1$, which is an ellipse crossing the $x$-axis at $\pm \frac{1}{4}$, and the $y$-axis at $\pm \frac{1}{2}$.

• I do not know what eigenvectors are and I don't see how the matrix correlates to the description of the ellipse. Jan 17, 2013 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 $aX^2+bXY+cY^2$. Jan 18, 2013 at 18:57
• @FlybyNight What about vice versa of the question? May 11, 2016 at 7:44
• Could you please explain the above concept with an example, such as $(10x^2 + 10y^2 + 12xy = 1)$. Feb 7, 2017 at 14:29
• @AnshKumar I've added an example to my post. Hope it helps. Please remember to upvote answer that you find useful. Feb 8, 2017 at 16:35

$$Ax^2+Bxy+Cy^2+Dx + Ey +F=0 \hspace{2 mm} .. (1)$$ 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.

• 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. Jan 17, 2013 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. Jan 17, 2013 at 23:24
• it it explained here , sorry i had been writing it oppositely. fixed it Jan 17, 2013 at 23:33
• What do $X$ and $Y$ equal? Jan 18, 2013 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 Jan 18, 2013 at 11:28

For the vice versa part of the question: the parameterization for the ellipse centered at the origin and with semi major axis equal to $$a$$ (parallel to the frame of reference) and semi minor axis equal to $$b$$ is

$$\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}a\cos\varphi\\b\sin\varphi\end{pmatrix}\qquad\varphi\in[0,2\pi)$$

If I rotate the ellipse of the angle $$\theta$$ with the rotation matrix $$R=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$$ I will get:

$$\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\cdot\begin{pmatrix}a\cos\varphi\\b\sin\varphi\end{pmatrix}$$

If I translate the center from $$(0,0)$$ to $$(x_c,y_c)$$ I will get:

$$\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\cdot\begin{pmatrix}a\cos\varphi\\b\sin\varphi\end{pmatrix}+\begin{pmatrix}x_c\\y_c\end{pmatrix}$$

Now I would like to isolate $$\cos\varphi$$ and $$\sin\varphi$$, then square them and finally get the quadratic form.

$$\begin{pmatrix}x-x_c\\y-y_c\end{pmatrix}=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\cdot\begin{pmatrix}a\cos\varphi\\b\sin\varphi\end{pmatrix}$$

Since $$R$$ is a rotation matrix, it holds $$R^{-1}=R^T$$ and so I can multiply both sides by $$R^T$$:

$$\begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix}\cdot\begin{pmatrix}x-x_c\\y-y_c\end{pmatrix}=\begin{pmatrix}a\cos\varphi\\b\sin\varphi\end{pmatrix}$$

Now I do the matrix vector product:

$$\cos(\theta)(x-x_c)+\sin(\theta)(y-y_c)=a\cos\varphi$$

$$-\sin(\theta)(x-x_c)+\cos(\theta)(y-y_c)=b\sin\varphi$$

and I isolate $$\cos\varphi$$ and $$\sin\varphi$$:

$$\frac{\cos(\theta)(x-x_c)+\sin(\theta)(y-y_c)}{a}=\cos\varphi$$

$$\frac{-\sin(\theta)(x-x_c)+\cos(\theta)(y-y_c)}{b}=\sin\varphi$$

Now, since $$\cos^2\varphi+\sin^2\varphi=1$$, I get

$$\left(\frac{\cos(\theta)(x-x_c)+\sin(\theta)(y-y_c)}{a}\right)^2+ \left(\frac{\cos(\theta)(y-y_c)-\sin(\theta)(x-x_c)}{b}\right)^2=1$$

I will now use sympy to extract the coefficients $$A$$, $$B$$, $$C$$, $$D$$, $$E$$, $$F$$ of the quadratic form $$Ax^2 + By^2 + Cxy + Dx + Ey + F = 0$$

from sympy import *
theta,x_c,y_c,a,b,x,y=symbols('theta x_c y_c a b x y')
exp_1=((cos(theta)*(x-x_c)+sin(theta)*(y-y_c))/a)**2+((cos(theta)*(y-y_c)-sin(theta)*(x-x_c))/b)**2-1
exp_2=expand(exp_1)
P=poly(exp_2,x,y)
A,B,C,D,E,F=symbols('A B C D E F')
A_=P.coeff_monomial(x**2)
B_=P.coeff_monomial(y**2)
C_=P.coeff_monomial(x*y)
D_=P.coeff_monomial(x)
E_=P.coeff_monomial(y)
F_=P.coeff_monomial(1)
print(latex(relational.Eq(A, A_)))
print(latex(relational.Eq(B, B_)))
print(latex(relational.Eq(C, C_)))
print(latex(relational.Eq(D, D_)))
print(latex(relational.Eq(E, E_)))
print(latex(relational.Eq(F, F_)))


$$A = \frac{1}{b^{2}} \sin^{2}{\left (\theta \right )} + \frac{1}{a^{2}} \cos^{2}{\left (\theta \right )}$$ $$B = \frac{1}{b^{2}} \cos^{2}{\left (\theta \right )} + \frac{1}{a^{2}} \sin^{2}{\left (\theta \right )}$$ $$C = - \frac{2}{b^{2}} \sin{\left (\theta \right )} \cos{\left (\theta \right )} + \frac{2}{a^{2}} \sin{\left (\theta \right )} \cos{\left (\theta \right )}$$ $$D = - \frac{2 x_{c}}{b^{2}} \sin^{2}{\left (\theta \right )} + \frac{2 y_{c}}{b^{2}} \sin{\left (\theta \right )} \cos{\left (\theta \right )} - \frac{2 x_{c}}{a^{2}} \cos^{2}{\left (\theta \right )} - \frac{2 y_{c}}{a^{2}} \sin{\left (\theta \right )} \cos{\left (\theta \right )}$$ $$E = \frac{2 x_{c}}{b^{2}} \sin{\left (\theta \right )} \cos{\left (\theta \right )} - \frac{2 y_{c}}{b^{2}} \cos^{2}{\left (\theta \right )} - \frac{2 x_{c}}{a^{2}} \sin{\left (\theta \right )} \cos{\left (\theta \right )} - \frac{2 y_{c}}{a^{2}} \sin^{2}{\left (\theta \right )}$$ $$F = -1 + \frac{x_{c}^{2}}{b^{2}} \sin^{2}{\left (\theta \right )} - \frac{2 x_{c}}{b^{2}} y_{c} \sin{\left (\theta \right )} \cos{\left (\theta \right )} + \frac{y_{c}^{2}}{b^{2}} \cos^{2}{\left (\theta \right )} + \frac{x_{c}^{2}}{a^{2}} \cos^{2}{\left (\theta \right )} + \frac{2 x_{c}}{a^{2}} y_{c} \sin{\left (\theta \right )} \cos{\left (\theta \right )} + \frac{y_{c}^{2}}{a^{2}} \sin^{2}{\left (\theta \right )}$$

• This is good work. I wanna ask what is the following formula called? Is it called canonical form of the rotated and translated ellipse? $$\left(\frac{\cos(\theta)(x-x_c)+\sin(\theta)(y-y_c)}{a}\right)^2+ \left(\frac{\cos(\theta)(y-y_c)-\sin(\theta)(x-x_c)}{b}\right)^2=1$$ Nov 29, 2020 at 16:02
• @GENIVI-LEARNER I am sorry, I do not know. Nov 30, 2020 at 7:48