# Orthogonality of major and minor axes of an ellipse

How would one go about proving that the major and minor axes of an ellipse are perpendicular bisectors of each other? I've read several proofs of expressions for the ellipse in cartesian and polar coordinates given the definition of the ellipse as the collection of points where the sum of the distance from two points is constant, but they always just assume that it is true. Visually it looks true, but is there an argument that follows simply from the definition?

• What are your definitions of major and minor axes? I guess they are not "the line through the foci is called the major axis, and the line perpendicular to it through the center is called the minor axis" because then the statement would be true by definition and you wouldn't ask. So, what definitions should we use? – Kamil Maciorowski Dec 8 '17 at 19:30
• I was under the impression that the major and minor axes names derived from their size as it was apparent to me that the longest line passing through the ellipse was the major axis. At the time of writing the original question I assumed that the minor axis was defined as being the shortest distance going through the center and two points on the ellipse. I saw it illustrated, but no reasoning behind it. Aretino's answer showed that the line perpindicular to the major axis, the minor axis under the standard definition, indeed is the shortest line going through the center. – Qsdd Dec 11 '17 at 15:25

If $SS'=2c$ is distance between the ellipse foci, and $2a=PS+PS'$ the sum of the distances between any point $P$ of the ellipse and the foci, then it is not difficult to show that the distance between $P$ and the center $O$ of the ellipse can be written as $$OP^2={c^2\over a^2}x^2+a^2-c^2,$$ where $x$ is the distance between $O$ and the projection of $P$ onto line $SS'$. From that relation it follows that you have the smallest distance $OP$ for $x=0$, while the largest distance is obtained when $P$ lies on $SS'$ and $x=a$.

EDIT.

Here's a possible derivation of the above equation.

Let $H$ be the projection of point $P$ on line $SS'$ and define: $x=OH$, $y=PH$, $d=PS$. Applying Pythagora's theorem to triangles $PHS$ and $PHS'$ we have: \begin{align} (c-x)^2+y^2 & =d^2\\ (c+x)^2+y^2 & =(2a-d)^2 \end{align} Subtracting these equations we get $\displaystyle d=a-{c\over a}x$ and plugging this into the first equation we obtain $$OP^2=x^2+y^2={c^2\over a^2}x^2+a^2-c^2.$$

• I follow your argumentation when the form for OP is found, but I've been tooling around with this for an hour, and have yet to find your formula. I may just be missing something elementary. You construct a right angled triangle from the projection, correct? How did you find the expression for the leg which does not have length x? – Qsdd Dec 8 '17 at 14:09
• See my edited answer. – Aretino Dec 8 '17 at 14:38

As a very simple argument you can consider that an ellipse could be obtained by stretching a circle along a diameter, thus major and minor axes of an ellipse are perpendicular. The reason for this is very clear considering the canonical form of an ellipse: $$\frac{x^2}{a^2}+ \frac{y^2}{b^2}=1$$ which can be transformed in a circle stretching one axis.

EG $$z=\frac{b}{a} y \to x^2+ z^2=a^2$$

This is also the reason for the formula to calculate the area of the ellipse $A=\pi ab$.

• But that seems to be a different definition of "ellipse" from the one based on the sum of the distance from the foci being constant, so you have to show that the two definitions are equivalent. Which doesn't seem any easier than the original problem. – Michael Kay Dec 8 '17 at 16:35
• @MichaelKay I've added an explanation directly in the answer. Thanks. – user Dec 8 '17 at 17:02

As per the principal axis theorem, the principal axes are given by the eigenvectors of the quadratic form that defines the ellipse. This quadratic form is symmetric, and therefore (by the spectral theorem) its eigenvectors are orthogonal. $\blacksquare$

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Naturally, we assume that the eccentricity is non-zero; otherwise, the claim is vacuous.

• Note also: this argument trivially generalises to higher dimensions. – AccidentalFourierTransform Dec 9 '17 at 9:57

The synthetic definiton of an ellipse is the geometric locus of all points such that the sum of the distances from the foci is constant. By a symmetry argument, it follows that the locus is symmetric with respect to the line passing through the foci and with respect to the axis of the segment connecting the two foci. The major and minor axes are defined to be segments of these lines, so they are perpendicular one to the other.

• I had in mind that the minor axis was the smallest distance across the ellipse. That the major axis as defined here is the longest distance is apparent, but I'm still having trouble following the argument for the minor axis as defined being the smallest length. – Qsdd Dec 8 '17 at 12:09
• Oh. I always knew that the axes were actually the axes of symmetry, and in an ellipse there are two of them, in general of different lenghts, hence the distinction between "miinor" and "major". – marco trevi Dec 8 '17 at 12:23