How would you prove/explain that an ellipse of equation x^2/9 + y^2/4 = 1 is inscribed by a rhombus which has sides tangent to the ellipse.
This is a
great question. Of course, this would never be asked in an exam since this is more of an investigation question. Had to dig pretty deep for this one. One way to go about your question is to just try to find coordinates of a rhombus for which the ellipse is inscribed in. But this could be tedious, and so it might be wise to prove that for ANY ellipse, you could find a rhombus for which that ellipse is inscribed.
First, let's start with a circle of radius \(1\) centered at the origin. It should be pretty easy to see that we could form a square using 4 points on the coordinate axes whose sides are tangential to the circle. That is, let \[\mathcal{C}:\ x^2+y^2=1\] and let \(\mathcal{S}\) be a square with vertices at \((0,\,\pm p)\) and \((\pm p,\,0)\), where \(p>0\).
It is clear by symmetry that the line segment joining the points \((0,\,p)\) and \((p,\,0)\) is tangential to \(\mathcal{C}\) at the point \(\left(\dfrac{1}{\sqrt{2}},\,\dfrac{1}{\sqrt{2}}\right)\), and so by equating gradients we have \[\frac{p-1/\sqrt{2}}{0-1/\sqrt{2}}=-1\implies p=\sqrt{2}.\] Now, we leverage the fact that properties like tangency are preserved under dilations from the coordinate axes.
That is, the ellipse given by \[\mathcal{E}:\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\quad a,b>0\] can be obtained from the circle \(\mathcal{C}\) via \begin{align*}&1.\ \ \text{a dilation by factor }a\text{ from the }x\text{-axis}\\
&2.\ \ \text{a dilation by factor }b\text{ from the }y\text{ -axis}, \end{align*} and so the parallelogram \(\mathcal{R}\) with vertices at \((\pm a\sqrt{2},\,0)\), \((0,\,\pm b\sqrt{2})\) will necessarily inscribe the ellipse \(\mathcal{E}\).
All that remains is to show that \(\mathcal{R}\) has equal side lengths to prove that the parallelogram we formed is indeed a rhombus, which is a pretty trivial task, since the expressions for the length of any of its sides is identical: \[d=\sqrt{(a\sqrt{2})^2+(b\sqrt{2})^2}=\sqrt{2a^2+2b^2}.\]
So, going back to the original problem we can say that the ellipse given by \(\dfrac{x^2}{3^2}+\dfrac{y^2}{2^2}=1\) is inscribed by a rhombus with vertices at \[(\pm 3\sqrt{2},\,0)\ \ \text{and}\ \ (0,\,\pm 2\sqrt{2}).\]
Going to finish off this post with a quick note: This is by
no means the most general result. In fact, it can be shown that the are
infinitely many rhombi that could inscribe
any ellipse.