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    Find the circumscribed square an ellipse x^2/a^2 + y^2/b^2 = 1 so that:
    - the square has smallest area
    - the square has greatest area.
    (find the equations for all the sides of the square)
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    (Original post by BCHL85)
    Find the circumscribed square an ellipse x^2/a^2 + y^2/b^2 = 1 so that:
    - the square has smallest area
    - the square has greatest area.
    (find the equations for all the sides of the square)
    I'll give you an idea of how to do this, then pick it up from there:

    Take only positive y values of the ellipse, so that we have the upper half. The equation of this upper half is y = sqrt[(b^2)(1 - (x/a)^2)].

    Now consider only one quadrant of the ellipse. The area of this one quadrant will obviously be yx (area of square = height x base -> in this case height is the y axis, and the base is the x axis). Hence the entire area of the square/rectangle inscribed in the ellipse is 4 times the area of one quadrant.

    So entire area = 4xy = 4x*sqrt[(b^2)(1 - (x/a)^2)]. Differentiate this equation and set it equal to 0 to find the maximum/minimum. To check which value of x is the maximum area differentiate 4xy twice and note that the solution should be < 0. Similarly for the minimum the solution should be > 0.

    Plug in those values of x into the equation 4x*sqrt[(b^2)(1 - (x/a)^2)] and you will have the respective maximum and minimum areas of the inscribed square in that ellipse.

    Euclid.
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    (Original post by Euclid)
    I'll give you an idea of how to do this, then pick it up from there:

    Take only positive y values of the ellipse, so that we have the upper half. The equation of this upper half is y = sqrt[(b^2)(1 - (x/a)^2)].

    Now consider only one quadrant of the ellipse. The area of this one quadrant will obviously be yx (area of square = height x base -> in this case height is the y axis, and the base is the x axis). Hence the entire area of the square/rectangle inscribed in the ellipse is 4 times the area of one quadrant.

    So entire area = 4xy = 4x*sqrt[(b^2)(1 - (x/a)^2)]. Differentiate this equation and set it equal to 0 to find the maximum/minimum. To check which value of x is the maximum area differentiate 4xy twice and note that the solution should be < 0. Similarly for the minimum the solution should be > 0.

    Plug in those values of x into the equation 4x*sqrt[(b^2)(1 - (x/a)^2)] and you will have the respective maximum and minimum areas of the inscribed square in that ellipse.

    Euclid.
    Hmm, I think what wrote made u misunderstand . It's a circumscribed square of an ellipse means an ellipse is inscribed in the square. Sorry about that.
    However, I did read ur hint, but I didn't find how to find a square, not rectangle. I think u were trying to find a rectangle? :confused:
    Anyway, thanks.
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    (Original post by BCHL85)
    Hmm, I think what wrote made u misunderstand . It's a circumscribed square of an ellipse means an ellipse is inscribed in the square. Sorry about that.
    However, I did read ur hint, but I didn't find how to find a square, not rectangle. I think u were trying to find a rectangle? :confused:
    Anyway, thanks.
    Ah right, I thought you meant find the maximum area of the four sided shape with parallel opposite sides inscribed in that ellipse you gave the equation of, sorry .

    To answer your question, look at it this way:

    The square will touch the ellipse at the points where x = +/- a and y = +/- b.
    Then integrate the equation y = sqrt[(b^2)(1 - (x/a)^2)] between limits a and -a to get the the area of the upper side of the ellipse. Multiply by 2 to get the entire area of the ellipse.

    Euclid.
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    It's ok. But the problem is still there without any hint
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    (Original post by Euclid)
    The square will touch the ellipse at the points where x = +/- a and y = +/- b.

    Why need this be the case/ Why couldn't the square be drawn at an angle, rather than parallel to the axes?
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    (Original post by RichE)
    Why need this be the case/ Why couldn't the square be drawn at an angle, rather than parallel to the axes?
    Because (I think!) you would find that would be the maximum area.

    Euclid.
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    (Original post by Euclid)
    Because (I think!) you would find that would be the maximum area.

    Euclid.
    Then what does circumscribed mean in this case? Does it only have to meet at two points, the way such a square would, rather than four?
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    (Original post by RichE)
    Then what does circumscribed mean in this case? Does it only have to meet at two points, the way such a square would rather than four?
    No it can meet at four points. Do it diagramatically and you will see. Although the points of intersection I suggested earlier could be incorrect, as it may have suggested that if the shape is a square => a=b.

    Euclid.
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    (Original post by Euclid)
    No it can meet at four points. Do it diagramatically and you will see. Although the points of intersection I suggested earlier could be incorrect, as it may have suggested that if the shape is a square => a=b.

    Euclid.
    Right, it can meet at 4 points, but is it the unique circumscribed square? Let's try with rectangle circumscribes the ellipse. I think we should find the equation of the tangent to the ellipse?
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    (Original post by BCHL85)
    Right, it can meet at 4 points, but is it the unique circumscribed square? Let's try with rectangle circumscribes the ellipse. I think we should find the equation of the tangent to the ellipse?
    Yes I think the square would be unique. Infact the ellipse inside it would only form two different angles in order to touch all four sides.

    The equation of the tangent to the ellipse at (acost,bsint) is:
    aysint + bxcost = b.a

    Euclid.
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    (Original post by Euclid)
    The equation of the tangent to the ellipse at (acost,bsint) is:
    aysint + bxcost = b.a

    Euclid.
    We have bcost = A, asint = B, ab = C
    A^2 + B^2 = (asint)^2 + (bcost)^2 = (ab)^2 = C^2.
    Ax + By + C = 0 is the tangent of the ellipse if A^2 + B^2 = C^2.
    So we got 2 parallel tangents are:
    Ax + By +/- C = 0.
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    Ah, then we find the equations for the 2 other tangents which are perpendicular to the two above : Bx - Ay +/- C = 0. (still applied B^2 + (-A)^2 = C^2 -> tangent?). Is it right?
 
 
 
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