# find the minimum area of rectangle outside a specific ellipse

I've found a similar question solving by calculus.

Find the area of largest rectangle that can be inscribed in an ellipse

i just want to prove the rectangle with minimum area must has the same symmetry axis with the ellipse($\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$). Then i can easily get the answer : $4ab$.

is it true for any closed curve with symmetry? for example $((x-2)^2+y^2)\cdot((x+2)^2+y^2)=256$

• But is the rectangle outside the ellipse, or inside? – Aretino Apr 26 '17 at 14:27
• @Aretino outside – Charles Bao Apr 27 '17 at 0:34

In an ellipse, all circumscribed parallelograms whose sides are parallel to a pair of conjugate diameters have the same area. And it is easy to show that any other parallelogram circumscribed to the same ellipse has a greater area.

Proof. Consider (diagram below) parallelogram $EFGH$, whose sides are tangent to the ellipse and parallel to conjugate diameters $AB$ and $CD$. Take any other diameter $PP'$: tangents at its endpoints form with lines $FG$, $HE$ another parallelogram $IJKL$, circumscribed to the ellipse. As line $IJ$ cuts $EF$ at $N$, the difference between the areas of $IJKL$ and $EFGH$ is equal to twice the difference between the areas of triangles $JFN$ and $IEN$. But $C$ is the midpoint of $EF$ and $N\ne C$, so that outer triangle $JFN$ is larger than inner triangle $IEN$. It follows that parallelogram $IJKL$ is larger than $EFGH$.

In particular, among all parallelograms whose sides are parallel to a pair of conjugate diameters, we can consider the rectangle whose sides are parallel to the ellipse axes of symmetry: by the above proof, it is the smallest rectangle circumscribed to the ellipse.

But it is not true, in general, that if a closed curve has two orthogonal axes of symmetry, then the circumscribed rectangle with minimum area has the same symmetry. Consider for instance the curve formed by the three circles in the diagram below. Red rectangle has area $4a(2\sqrt{ab}+b)$ while blue rectangle has a larger area of $4a(a+2b)$.

If you make the linear transformation $(x',y')=(x/a,y/b)$ the ellipse becomes a circle of radius 1 and a rectangle a rhomb.

Areas transforms by a constant. The rhomb has height 2 (a diameter in the circle) and base $\geq 2$ with equality iff the rhomb is a square iff the angles of the original rectangle are preserved under the above linear transformation, and this happes precisely if it is aligned along the axis as you suggested.

For the general case, I think the question is a bit too vague.

Since 4 vertexes of rectangle is on a circle, if you know the normal case, you see first rectangle which is parallel to x and y axis is minimum.