[RESOLVED] calculate largest rectangle that fits in ellipse

how can i calculate the largest rectangle that fits in an ellipse + the 4 points that describe that rectangle?
see image... the green rectangle is what i'm looking for:

Attachment 91973

The ellipse in your case is given by the equation

x^2 / A^2 + y^2 / B^2 = 1

where 2A is the width and 2B is the height of the ellipse. You wish to find a point (p, q) on the ellipse such that the area pq is maximized. Solve the above equation for y to get an expression valid for y >= 0:

y^2 / B^2 = 1 - x^2 / A^2
y = B sqrt(1 - x^2 / A^2)

The area pq is then just p*B sqrt(1 - p^2 / A^2). This is maximized when the derivative is zero, so differentiate with respect to p:

(pq)' = B sqrt(1 - p^2 / A^2) + B p*1/2*(1 - p^2 / A^2)^(-1/2) * (-2p/A^2) = 0
=> (1 - p^2 / A^2) - p^2 / A^2 = 0
=> 1 = 2p^2 / A^2 and p^2 / A^2 = 1/2
=> p = A sqrt(2)/2
=> q = B sqrt(1 - 1/2) = B sqrt(2)/2

The four corners are then just (+/- p, +/- q). The maximal area is 4pq = 2AB. The general case of a rotated, translated ellipse can be reduced to this case: take the center of the ellipse C, the axis X with constant A, the axis Y with constant B, and just move A*sqrt(2)/2 along the X axis or B*sqrt(2)/2 along the Y axis to get to your points. Computing these axes is another problem and depends on the form in which the ellipse is given. For the quadratic form, it's just completing the square.


Alternative derivation:
Consider a circle of radius 1. The area is maximized when p=q from symmetry and intuition. This occurs for points (p, q) = (sqrt(2)/2, sqrt(2)/2). Imagine scaling the vertical axis by B. Two rectangles pre-scaling have their area increased by a factor of B by the scaling, so the maximal-area rectangle is the same before and after scaling. Do the same with the horizontal axis, multiplying by A. The resulting points (p, q) are just (A sqrt(2)/2, B sqrt(2)/2), as I derived more rigorously from calculus above.

i resolved it by calculating height * (4/3) and width * (4/3)
it's a close enough approximation for my purposes

thanks for answering

Sure, though as my derivation shows, using height * sqrt(2) and width * sqrt(2) is exact. Yours is a decent approximation since

sqrt(2) = 1.4142...
4/3 = 1.3333...

are reasonably close.