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Math Help - Minimize Area of ellipsis

  1. #1
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    Minimize Area of ellipsis

    Consider an ellipse of equation \frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1 that contains the circle x^2+y^2=2y

    The question is: Wich values of a and b minimize the area of the ellipse?
    Last edited by orbit; May 19th 2009 at 11:35 AM.
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  2. #2
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    Quote Originally Posted by orbit View Post
    Consider an ellipsis of equation \frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1 that contains the circle x^2+y^2=2y

    The question is: Wich values of a and b minimize the area of the ellpsis?
    The area of the ellipse \frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1 is \pi ab. Now, what is the circle x^2+ y^2= 2y? That is, where is its center and what is its radius? It looks to me like you want your ellipse to just contain that circle which means it must be tangent (at, I think, two points).
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  3. #3
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by orbit View Post
    Consider an ellipse of equation \frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1 that contains the circle x^2+y^2=2y

    The question is: Wich values of a and b minimize the area of the ellipse?
    Hi orbit.

    My guess is that the area of the ellipse is minimized when it’s a circle.
    Last edited by TheAbstractionist; May 19th 2009 at 01:09 PM.
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  4. #4
    Senior Member TheAbstractionist's Avatar
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    A circle passing through (0,2) so a=b=2.
    Last edited by TheAbstractionist; May 19th 2009 at 01:11 PM.
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  5. #5
    Junior Member rubix's Avatar
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    like HallsofIvy said, "you want your ellipse to just contain that circle which means it must be tangent". In other words, they'll have the same dy/dx. This suggests you to take the derivative.

    Futhermore, you also need to realize that at that point y value shall be the same.

    So, solve x^2 in circlue equation in terms of y.

    Then, plug that value in x^2 of the ellispe equation.

    Now write the ellipse equation for y in terms of a and b (remember you already eliminated x by substitution).

    Next, use implicit differentiation on both equations.


    edit: that's an excellent idea TheAbstractionist gave you. Should be lot faster than what i said
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  6. #6
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    Quote Originally Posted by TheAbstractionist View Post
    A circle passing through (0,2) so a=b=2.
    I thought the same thing, but I donīt know if it`s alright. Because if we think the ellipse as the circle, the problem is too easy.

    Lol
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  7. #7
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    Anybody else wants to say something?
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