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?
 A: 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 $A=\pi ab$.
A: 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
$$
OP^2=x^2+y^2={c^2\over a^2}x^2+a^2-c^2.
$$
A: 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. $\blacksquare$
--
Naturally, we assume that the eccentricity is non-zero; otherwise, the claim is vacuous.
A: 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.
