<s>[INST] How to show that any rectangle in ellipse must be oriented parallel to axes? A problem which is often given as an exercise for students learning about calculus and finding extrema, is to find maximal possible area of a rectangle inside an ellipse. Such question was asked, for example, here: Find the area of largest rectangle that can be inscribed in an ellipse (A similar problem in three dimensions is also often asked: Dimensions of a box of maximum volume inside an ellipsoid.)
The solution usually starts by stating that we the rectangle must be oriented in such way that the sides are parallel to the ellipse, which gives a simple expression for the area.
Even though the fact that any rectangle inscribed in an ellipse must be oriented in this way seems intuitively clear, I would like to see an argument showing that this is indeed the case. (I have posted as an answer my attempt using analytic geometry.)
Of course, there is one obvious exception. Rectangle inscribed in a circle can be rotated in any direction we want. So we should assume that the semiaxes of the ellipse have different lengths.
Here is a picture illustrating the situation (shamelessly stolen from this post):

 [/INST] Suppose we have a rectangle which has all four vertices on the ellipse. If $C$ is the center of the rectangle, then all vertices of the rectangle are in the same distance from $C$. If we choose coordinate system with origin in the point $C$, we get the equation
$$x^2+y^2=r^2.$$
We also want to express the condition that the vertices belong to the ellipse. So far we have only chosen the origin of our Cartesian coordinate system to be the point $C$. We can also choose the axes to be parallel to the axes to the ellipse. If we denote by $(x_0,y_0)$ the center of the ellipse, we get the equation
$$\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1.$$
Since rectangle is centrally symmetric w.r.t. the point $C=(0,0)$, the vertices also lie on an ellipse which is centrally symmetric to the ellipse described above. This si precisely the ellipse
$$\frac{(x+x_0)^2}{a^2}+\frac{(y+y_0)^2}{b^2}=1.$$
So we got the following three equations:
\begin{align}
x^2+y^2&=r^2 \tag{1}\\
b^2(x-x_0)^2+a^2(y-y_0)^2&=a^2b^2 \tag{2}\\
b^2(x+x_0)^2+a^2(y+y_0)^2&=a^2b^2 \tag{3}
\end{align}
We are asking for the conditions on $x_0$, $y_0$ and $r$ such that there are at least four solutions.
Notice that by subtracting the equations $(2)$ and $(3)$ we get
$$b^2xx_0+a^2yy_0=0.\tag{4}$$
Let us first show that $x_0=y_0=0$. Assume that this is not true. For example, let us assume $x_0\ne0$. Then we get
$$x=\frac{a^2y_0}{b^2x_0}\cdot y. \tag{5}$$
By plugging $(5)$ into $(1)$ we get 
$$y^2\left(1+\frac{a^2y_0}{b^2x_0}\right)^2=r^2 \tag{6}$$
which gives us two possibilities for $y$. Combining this with $(5)$ we get two possibilities for the point $(x,y)$.
So we see that if $x_0\ne0$, then there are at most two solutions of the above system. The case $y_0\ne0$ is symmetric. In order to get a rectangle we need at four points, so we necessarily have1
$$x_0=y_0=0.\tag{7}$$
In the other words, we have found out that the center of the rectangle and the center of the ellipse coincide.
This condition significantly simplifies the original system to
\begin{align}
x^2+y^2&=r^2 \\
b^2x^2+a^2y^2&=a^2b^2 
\end{align}
From the first equation we get $y^2=r^2-x^2$. By plugging this value into the second equation we get
\begin{align}
b^2x^2+a^2(r^2-x^2)&=a^2b^2\\ 
(b^2-a^2)x^2&=a^2(b^2-r^2)
\end{align}
Since we do not consider the case $b=a$, which corresponds to the ellipse being circle, we get
$$x^2=\frac{a^2(b^2-r^2)}{b^2-a^2}.$$
Assuming the RHS is positive, this gives us two possible values $$x=\pm\sqrt{\frac{a^2(b^2-r^2)}{b^2-a^2}}.$$
From these values we can get $$y=\pm\sqrt{r^2-x^2}$$
(again assuming that $r^2-x^2>0$).
Clearly, we got a rectangle with the sides parallel to the axes. (The vertices of rectangle have the form $(x_1,y_1)$, $(-x_1,y_1)$, $(-x_1,y_1)$, $(-x_1,-y_1)$.)

In fact, using the above equations we can also find the conditions on $a$, $b$ and $r$ when the intersection exists. W.l.o.g. we can assume $0<a<b$. (The other case is symmetric.) If we want $a^2(b^2-r^2)/(b^2-a^2)$ to be positive, we need $r<b$.
We also need 
$$0<r^2-x^2 = r^2- \frac{a^2(b^2-r^2)}{b^2-a^2} = \frac{a^2(b^2-r^2)}{b^2-a^2} = \frac{b^2(r^2-a^2)}{b^2-a^2}$$
which gives us $a<r$.
So we get a solution only if
$$a<r<b$$
as expected. 

1Another possibility how to view the case $(x_0,y_0)\ne(0,0)$, i.e. the case when the center of the rectangle is different from the center of the ellipse, is that in such case we can only get a line segment - which could be considered as a "rectangle having zero width".
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