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Ellipse And Minimum Parallelogram (Posted on 2008-01-04) Difficulty: 3 of 5
Determine the minimum area of a parallelogram each of whose four sides are tangent to the ellipse x2/49 + y2/36 = 1.

See The Solution Submitted by K Sengupta    
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Solution Solution | Comment 2 of 4 |

Apply a compression transform aong the x-axis.  The ellipse is then transformed to a circle. 

Take the fact that the quadrilateral of minimum area around a circle is a square with sides equal to the circle's diameter.  The square can be in any orientation, but for simplicity, take the orientation with sides parallel to the axis.

Now invert the transform on the square and circle.  Since the compression transform preserves tangency, the rectangle formed from the square is tangent to the ellipse.  For the given ellipse, the rectangle has dimensions 12x14 and area 168.

If the diagonals of the square were parallel to the axis, then the resulting quadrilateral would be a rhombus with diagonals of 12*sqrt(2) and 14*sqrt(2).  The area of the rhombus is the same as the rectangle: 168.


  Posted by Brian Smith on 2008-01-05 23:11:54
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