Calculating area of square within an ellipse

**RedstarGuy** · April 12th 2008, 05:13 PM

Have a home work problem I'm not too sure about. Need to know what the correct method for calculating the area and generic equation for a square area within an ellipse.

Can anybody out there point me in the right direction? Any help you may be willing to provide will be greatly appreciated.

Thanks

Redstarguy

**mr fantastic** · April 13th 2008, 05:20 AM

Originally Posted by RedstarGuy

Have a home work problem I'm not too sure about. Need to know what the correct method for calculating the area and generic equation for a square area within an ellipse.

Can anybody out there point me in the right direction? Any help you may be willing to provide will be greatly appreciated.

Thanks

Redstarguy

There are several approaches. One approach is the following:

Since area is invariant under rotation and translation, an ellipse of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ can be assumed WLOG. The problem is to find the area of (I assume) the largest square that can fit in this ellipse.

To get the coordinates of the corners of the square (and hence the length of a side and hence the area), solve

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ .... (1)

$x = y$ .... (2)

simultaneously:

$\frac{x^2}{a^2} + \frac{x^2}{b^2} = 1 \Rightarrow b^2 x^2 + a^2 x^2 = a^2 b^2 \Rightarrow x^2 = \frac{a^2 b^2}{a^2 + b^2} \Rightarrow x = \pm \frac{ab}{\sqrt{a^2 + b^2}}$ .

Hence the largest square has sidelength $\frac{2ab}{\sqrt{a^2 + b^2}}$ .

Hence the area of the largest square is $\frac{4a^2 b^2}{a^2 + b^2}$ .

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