Angle ψ Between the Ellipse Axis and the x-Axis [Classical Mechanics I Studied #27]

Back in my early college years…

Ellipse equation: x²/A² − xy·2cos Δ/(AB) + y²/B² = sin²Δ

Textbook figure: the ellipse inside its 2A by 2B box, with the major axis tilted by ψ from the x-axis

when there’s an ellipse like this, for the angle psi that the major axis makes with the x-axis (sweats), the classical mechanics textbook had written the relation like this,

$$\tan 2\psi = \frac{2AB\cos\Delta}{A^2 - B^2}$$

That fixes the axes, but tangent alone can swap the long and short ones! To select the major axis, use $\psi=\tfrac12\operatorname{atan2}(2AB\cos\Delta,A^2-B^2)$, modulo $\pi$. If $A=B$ and $\cos\Delta=0$, it’s a circle and there’s no unique major axis.

and left the derivation of this formula as homework. While moving that post over to English, I realized I hadn’t properly derived it back then. T_T

Well… anyway, today I did manage to do it.

The reason I’m bothering to put up a post is that a lot of people ask about this, but a post with the derivation doesn’t seem to come up in searches, so I’m putting one up…

Actually, I don’t think what I worked out is the standard approach, and I’m sure there has to be an easier method, but anyway, I did prove it. T_T

Sigh…

Handwritten note: rotating by −ψ makes the xy term fly away, so plug in x = x′cos ψ − y′sin ψ and y = x′sin ψ + y′cos ψ

Handwritten note: the substitution expanded and grouped into x′², y′² and x′y′ terms; the x′y′ bracket is the one to make 0

Handwritten note: setting the x′y′ coefficient to 0 gives tan 2ψ = 2AB cos Δ/(A² − B²)

Heads-up. The figures here are the original notes, slips and all. A few things to watch for:

In the second line of the substitution, the first fraction is $\frac{(x'\cos\psi - y'\sin\psi)^2}{A^2}$, with $A^2$ below (the line writes $A$); the next line has $A^2$ again, so $\tan 2\psi = \frac{2AB\cos\Delta}{A^2 - B^2}$ is right.

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