How to calculate the bounding square of an ellipse?

Question

I'm trying to find the formula of a bounding square of an ellipse. So in other words: I want a formula that gives the size (length of one side of a the square) that bounds a given ellipse with width $a$ and length $b$.

I know that the formula of an ellipse is given by: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ but I just cant figure out what the length of one side of a bounding square is.

Answer 1

Sketch as an auxiliary figure a square and an elongated ellipse that is tangent to the four sides of the square. From this figure we can learn the following: The axes of the ellipse are lying on the diagonals of the circumscribing square. For the given ellipse ${x^2\over a^2}+{y^2\over b^2}=1$ this means that the sides of the circumscribing square are certain $45^\circ$ lines $x\pm y={\rm const.}$ Presenting this ellipse in the form $$x(t)=a\cos t,\quad y(t)=b\sin t\qquad(0\leq t\leq2\pi)$$ this implies that one side of the square touches the ellipse at the point where the objective function $$\rho(t):=x(t)+y(t)$$ is maximal. Now we all know that the maximal value of $\rho(t)=a\cos t+b\sin t$ computes to $\sqrt{a^2+b^2}$. It follows that the side length $s$ of the circumscribing square is given by $$s=\sqrt{2}\>\sqrt{a^2+b^2}\ .$$

The following figure shows a geometric construction of the circumscribing square:

Answer 2

Note that

• for $y=0 \implies x=\pm a$

• for $x=0 \implies y=\pm b$

therefore the rectangle with sides $2a$ and $2b$ centered at the origin bounds the ellipse and then also the square with side $L=\max\{2a,2b\}$ centered at the origin bounds the ellipse.

Answer 3

Did you actually mean a square or would a rectangle, such as gimusi showed, be sufficient? If you really want a square then it has to fit the larger of the two distances: side length 2a or 2b, whichever is larger. If a rectangle then "2a by 2b".