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How to find the length of major and minor axis of ellipse given the length of two conjugate diameters and the angle between them?

I am aware about how to construct the ellipse using the above given facts(not by Rytz's Construction). I would like to know, independent of what method of construction one uses, how one can find the length of the axes.

Aiytan
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2 Answers2

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Here's a geometric construction: if $MN$ and $DE$ are conjugate diameters, draw line $QQ'$ through $N$ perpendicular to $DE$ (see diagram below). Points $Q$ and $Q'$ must be chosen such that $NQ=NQ'=OD$. Major axis $IR$ is the bisector of angle $\angle QOQ'$ and minor axis $TS$ is perpendicular to it. Their lengths can be computed from: $$ \tag{1} IR=OQ'+OQ,\quad TS=OQ'-OQ. $$

enter image description here

If $ON=a$, $OD=b$ and the angle between them is $\theta$, then from the cosine rule applied to triangles $ONQ$ and $ONQ'$ we get: $$ OQ^2=a^2+b^2-2ab\sin\theta,\quad OQ'^2=a^2+b^2+2ab\sin\theta. $$ Inserting these into $(1)$ we finally obtain: $$ OR\cdot OS=ab\sin\theta,\quad OR^2+OS^2=a^2+b^2. $$ These equalities could have been directly derived, as they are well known properties of conjugate diameters (see properties 1. and 2. listed here).

Intelligenti pauca
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  • Hello, OP was asking for a construction-free method. – Parcly Taxel Aug 12 '19 at 18:12
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    @ParclyTaxel The equation in the last paragraph can be used independently of this construction, as has already been noted in the answer. These properties are “well known” enough to be mentioned in the Wikipedia article on ellipses. – amd Aug 12 '19 at 19:07
  • @amd Fine then, $a,b$ can be recovered by Viète. The SVD-based method is more general, however, since it also determines the ellipse orientation. – Parcly Taxel Aug 12 '19 at 19:10
  • @ParclyTaxel I wouldn’t characterize that as “general,” but as “more informative,” but that’s just a quibble. – amd Aug 12 '19 at 19:11
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Suppose the conjugate radii are $a$ and $b$ and the angle between them is $\theta$. Form the following matrix: $$\mathbf A=\begin{bmatrix}a&b\cos\theta\\ 0&b\sin\theta\end{bmatrix}$$ The columns are vectors corresponding to the conjugate radii. Now perform a singular value decomposition $\mathbf A=\mathbf{U\Sigma V}^T$. The diagonal entries of $\bf\Sigma$, the singular values of $\bf A$, are the semi-axis lengths.

This works because any ellipse centred on the origin is a linear transformation, $\bf A$ in this case, of the unit circle. The SVD corresponds to decomposing this transformation into a rotation/reflection $\mathbf V^T$ (which visually doesn't change anything), a scaling along the coordinate axes $\bf\Sigma$ (so that the ellipse semi-axes are its diagonal entries, as above) and another rotation/reflection $\bf U$ (which does not change the axis lengths). This is visualised below, with the displayed arrows being conjugate diameters of the resulting ellipse: In fact this method finds more than just the axis lengths. Suppose the columns of $\bf A$ represent any pair of conjugate radii vectors of an ellipse. Then the columns of $\bf U\Sigma$ are perpendicular conjugate radii, thus semi-axis vectors, for the same ellipse.

Parcly Taxel
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