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The traditional parametric equations of an ellipse centered at the origin are given by: $$ \{x(t)= a\cos(t)\cos(B) - b\sin(t)\sin(B); y(t)= a\cos(t)\sin(B) + b\sin(t)\cos(B)\} $$ where B is the angle of rotation of the major axis from the x-axis. Here, the major and minor axes lengths are given as a and b. However I have a different parametric form of an ellipse given by: $$ \{x(t)=A\cos(t) + B\sin(t); y(t)=C\cos(t) + D\sin(t)\}. $$ Each equation represents sinusoids along the the x and y axes with independent magnitudes and phases, thus forming an ellipse as the locus $(x(t),y(t))$.

The major and minor axes lengths are not apparent in this form. My approach has been to set the derivative $dx/dy = -(x/y)$ since the slope of the tangent is perpendicular to the radius only at the intersections with the major and minor axes. My result is $t=.5(\tan^{-1}((AB+CD)/2(A^2+C^2-B^2-D^2))$. This value of t does not quite match my numerical simulation (although it is close) and I am perplexed. Any help is greatly appreciated.

Thanks, Bob T.

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  • $\begingroup$ Are the $B$’s in the two parametrizations meant to be the same? $\endgroup$
    – amd
    Jan 29, 2018 at 20:39
  • $\begingroup$ Sorry, no I should have written these as separate variables $\endgroup$
    – Bob Tivnan
    Jan 29, 2018 at 20:54

3 Answers 3

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Hint:

From your equations, if they represents the same ellipse, we have: $$ A=a\cos \beta \qquad B=b\sin \beta $$ $$ C=a\sin \beta \qquad D=b\cos \beta $$ so: $$ \frac{C}{A}=\frac{B}{D}=\tan \beta $$

where $\beta$ is your angle $B$.

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  • $\begingroup$ But my A,B,C and D parameters are independent. So C/A does not necessarily equal B/D. I think part of the problem is that my equations use 4 parameters when only 3 are required for an ellipse. Wouldn't that mean that any given ellipse does not have a unique A,B,C,D combination and that your approach gives one such combination? Just spit balling here. $\endgroup$
    – Bob Tivnan
    Jan 29, 2018 at 21:47
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With Mathematica’s help, $t$ can be eliminated to get the implicit equation $$\left(C^2+D^2\right)x^2-2(AC+BD)xy=\left(AD-BC\right)^2-\left(A^2+B^2\right) y^2.$$ From here, you can use standard methods to find the principal axes and semi-axis lengths. The latter are eigenvectors of $${1\over(AD-BC)^2}\begin{bmatrix}C^2+D^2 & -(AC+BD) \\ -(AC+BD) & A^2+B^2\end{bmatrix},$$ i.e., $$\left[\left(A^2+B^2\right)-\left(C^2+D^2\right) \pm \sqrt{\left(A^2+B^2+C^2+D^2\right)^2-4\left(BC-AD\right)^2}, 2(AC+BD)\right]^T.$$ The semi-axis lengths are the reciprocal square roots of the corresponding eigenvalues.

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I found my mistake. I carelessly placed the 2 in the denominator of my equation:

t_axes=.5(arctan((AB+CD)/2(A^2+C^2^−D^2)) should be t_axes=.5(arctan(2(AB+CD)/(A^2+C^2−B^2−D^2))

This correctly gives the value of t that can be used to find the points of intersection of the principal axes with the ellipse. Thanks for the comments, which helped me get a new perspective on the problem.

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