The maximum area of a hexagon that can inscribed in an ellipseI have to find the maximum area of a hexagon that can inscribed in an ellipse. The ellipse has been given as x 2 16 + y 2 9 = 1.I considered the circle with the major axis of the ellipse as the diameter of the circle with centre as the origin. By considering lines of y = ± 3 x, you can find the points on the circle which are vertices of a regular hexagon(including the end points of the diameter). I brought these points on the given ellipse such that the x coordinate of the vertices remain the same and only the y coordinate changes depending upon the equation of the ellipse. But that did not lead to a hexagon.Since the question nowhere mentions of the hexagon being regular, I think it might be a irregular hexagon but any such hexagon can be drawn which encompasses most of the area of the ellipse.

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The maximum area of a hexagon that can inscribed in an ellipseI have to find the maximum area of a hexagon that can inscribed in an ellipse. The ellipse has been given as             x      2        16    +            y      2        9    =  1.I considered the circle with the major axis of the ellipse as the diameter of the circle with centre as the origin. By considering lines of   y  =  ±      3    x, you can find the points on the circle which are vertices of a regular hexagon(including the end points of the diameter). I brought these points on the given ellipse such that the x coordinate of the vertices remain the same and only the y coordinate changes depending upon the equation of the ellipse. But that did not lead to a hexagon.Since the question nowhere mentions of the hexagon being regular, I think it might be a irregular hexagon but any such hexagon can be drawn which encompasses most of the area of the ellipse.

Eve Good · Accepted Answer

Step 1We can use the affine transformation that   x  =  4  u  ,  y  =  3  v. Therefore, The ellipse will maps to the unit circle       u    2    +      v    2    =  1. One of the most important property of affine thansformations is invariance of ratio of the areas. Ratio of areas of ellipse and unit circle is   4  ⋅  3  =  12. Now we have to find a cyclic hexagon that has maximum area. (By isoperimetric inequaliy, this hexagon must be regular).Step 2Area of regular hexagon is                     3                  3                    2       and therefore, the hexagon has maximum area that in ellipse is   12  ⋅                    3                  3                    2        =  18      3  .

Dawson Downs · Accepted Answer

Step 1Consider the standard ellipse             x      2              a      2        +            y      2              b      2        =  1. An hexagon cab be inscribed in this ellipse whose area is 4 times the area of a rectangle plus a right angled triangle as       A    H    (  t  )  =  4  [  a  b  sin  ⁡  t  cos  ⁡  t  +      1    2    b  sin  ⁡  t  (  a  −  a  cos  ⁡  t  )  ]  =  a  b  [  sin  ⁡  2  t  +  2  sin  ⁡  t  ]                 (  1  ) where   P  (  a  cos  ⁡  t  ,  b  sin  ⁡  t  ) is the point on the ellipse and let the vertex be A(a,0) in the first quadrant. Now let us maximize       A    H    (  t  ), where             d              A        H                    d      t        =  0 gives   2  cos  ⁡  2  t  +  2  cos  ⁡  t  =  0    ⟹    t  =  π      /    3  ,  π.Step 2By choosing   t  =  π      /    3, we get       A    H    ″    (  π      /    3  )  &amp;lt;  0.. Finally,   m  a  x  (      A    H    )  =      A    H    (  π      /    3  )  =            3              3              2    a  b  =  18      3   (when   a  =  4  ,  b  =  3  )  .

Find the maximum area of a hexagon that can inscribed in an ellipse. The ellipse has been given as x^2/16+y^2/9=1.

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