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Math Help - Ellipse with rectangular frame

  1. #1
    Member courteous's Avatar
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    Question Ellipse with rectangular frame

    »Write down the equation of an ellipse, that goes through P(1,-2) and whose circumscribed* rectangle's area is minimized.«
    *

    (Pretending not knowing that "rectangle" is in fact square.)

    Set-up:
    Ellipse through (1,-2): \frac{1}{a^2}+\frac{4}{b^2}=1, rectangle's area: 2a*2b=min.
    Now, what to differentiate? From the first equation I've also got a^2b^2=b^2+4a^2.
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  2. #2
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    a^2b^2 = b^2 + 4a^2

    a^2b^2 - 4a^2 = b^2

    a^2(b^2 -4) = b^2

    a^2 = \frac{b^2}{b^2 - 4}

    a = \frac{b}{\sqrt{b^2 - 4}}

    A = 4ab

    A = \frac{4b^2}{\sqrt{b^2 - 4}}

    find \frac{dA}{db} and determine the value of b that minimizes the area.
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  3. #3
    MHF Contributor
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    Quote Originally Posted by courteous View Post
    »Write down the equation of an ellipse, that goes through P(1,-2) and whose circumscribed* rectangle's area is minimized.«
    *

    (Pretending not knowing that "rectangle" is in fact square.)

    Set-up:
    Ellipse through (1,-2): \frac{1}{a^2}+\frac{4}{b^2}=1, rectangle's area: 2a*2b=min.
    Now, what to differentiate? From the first equation I've also got a^2b^2=b^2+4a^2.
    You could use Lagrange multipliers. Minimizing A = 4ab subject to \frac{1}{a^2} + \frac{4}{b^2} = 1 is to use

    F = 4ab + \lambda \left( \frac{1}{a^2} + \frac{4}{b^2} - 1\right)

    then calculate F_a,\;\;\;F_b,\;\;\;F_{\lambda}, set them to zero and solve for a, b, \; \text{and}\; \lambda. It avoids the square roots. (You sure about the square?)
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