Equation of Rotated Ellipse - Semi Major Axis is Changing

Question

I am looking at astronomical observations of gas. The gas is orbiting a black hole with circular radius, $R$. However, from Earth it appears that the gas is an inclined ellipse. This is because projection effects cause the circle to be rotated once about its $y$ axis, creating an ellipse, before being rotated by an angle $\theta$ in the $z$ axis (out of the page). Perhaps the below image will help visualise this:

Image showing a circle being first rotated about the $y$ axis by an undefined amount, before being rotated about the $z$ axis by angle $\theta$.

I am looking to fit an ellipse to the $x,y$ coordinates I have for the gas on a map of the sky. From the general equation of an inclined ellipse, we have that,

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

and from how I have defined the rotation by $\theta$, we have that $b = R$. I can measure the semi-minor axis, $a$, on my map, however the gas emission does not extend out to the semi-major axis. I would like to determine $R$, hence I figured that if I rearranged the above for $b = R$, then I should find that each $x,y$ value would return the same value for $R$. Rearranging gives the following, $$ \frac{(x\sin(\theta)-y\cos(\theta))^2}{1-(\frac{x\cos(\theta)+y\sin(\theta)}{a})^2} = R^2,$$

at which point I should apologise if any errors have crept in at this point - the main point I'm trying to get across is that, to my reckoning, any $x,y$ on the gas ellipse will return the same value for $R$ (ignoring errors and physics) when plugged into the above equation. However, when I did do this, my values for $R^2$ decreased proportionally to $x^2 + y^2$. Again, let's totally ignore the physics here and focus on the mathematics. Can I please check if I have done anything wrong up to this point? Am I correct in stating that $b = R$ from the original circle, and that the above equation should hold for all points on the ellipse? What I also found odd, was that if I rearranged the first equation for $y$ by getting it in a quadratic form, and plotted $y$ vs $x$ on Desmos for different values of $\theta$, $a$ and $R$, I did not get an inclined ellipse. I would like to do this so I can plot different ellipses on my data and find the best fitting one. This equation for $y$ was pretty ghastly so I will spare you seeing it, but I assume my problem is the same as above, and I have made a mathematical error somewhere in my understanding.

Thank you!

I should add that my coordinate system is centered such that $x_0, y_0 = 0$

Answer 1

I am not sure what I am looking at, but I think for rotation it should be the sin that changes sign, not the cos.

Rotation matrix $$R_\alpha = \begin{bmatrix}\cos(\alpha)&-\sin(\alpha)\\\sin(\alpha)&\cos(\alpha)\end{bmatrix}$$

Answer 2

There is a great short video by “Senzen” and at the end of the video, a formula for your problem is shown. Video.

Assuming the center of the ellipse is at the origin, rotation by the angle θ for $ax^2+by^2=1$ would be: $a(y sinθ+x cosθ)^2+b(y cosθ-x sinθ)^2=1$.

Here is a demo: https://www.desmos.com/calculator/73tczm8buw.

To derive the formula, watch the video and solve for $y=\pm \sqrt {\frac {1-ax^2}{b}}$