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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.

Rodrigo de Azevedo
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Wouter61636
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    Isn't it that $a$ and $b$ are the radius in the given direction (a for x, b for y) – Finn Eggers Sep 24 '18 at 12:22
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    Do you need to handle just the case in standard orientation / configuration, or do you need to deal with a general 2nd order equation, which might describe an ellipse with rotation and/or translation? – mvw Sep 24 '18 at 12:34
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    Are you looking for the size of the smallest square that bounds the ellipse? – Barry Cipra Sep 24 '18 at 14:18
  • You’ll need to be more specific as there are many possible “bounding squares.” – amd Sep 24 '18 at 19:26
  • a and b are indeed the radius in given directions for an ellips. For instance: an ellips with a=3 and b=2 would be bounded by a rectangle of 32 or a square of 33 indeed. As mentioned by Barry Cipra, i am indeed looking for the smallest bounding square which is free to rotate in any direction relative to the ellips. – Wouter61636 Sep 25 '18 at 13:41

3 Answers3

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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.

enter image description here

user
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  • The OP has asked for a square, so the square would have side length of $2\max(a,b)$ – lioness99a Sep 24 '18 at 12:24
  • @lioness99a Yes you are right! My initial aim was only give a hin. tI've added also that consideration. – user Sep 24 '18 at 12:26
  • Thanks for your answer and thaughts and sorry that i wasn't clear in my initial question: i am looking for the smallest bounding square for the ellips. If you use 2max(a,b) you'll get a bounding square, but not the smallest. I am sorry i wasn't specific on this in my initial question. – Wouter61636 Sep 25 '18 at 13:44
  • Ok now I see what you are looking for! Bye – user Sep 25 '18 at 16:58
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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:

enter image description here

Christian Blatter
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  • Thank you! Could you maybe elaborate on how the maximal value of ρ(t)=acos(t)+bsin(t) computes to sqrt(2)*sqrt(a^2+b^2)? – Wouter61636 Sep 25 '18 at 14:04
  • @Wouter61636: See here: https://math.stackexchange.com/questions/560711/find-the-maximum-and-minimum-values-of-a-cos-t-b-sin-t – Christian Blatter Sep 25 '18 at 15:11
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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".

user247327
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