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Math Help - Find eq. to ellipse

  1. #1
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    Find eq. to ellipse

    Find the equation of the line that is tangent to the ellipse
    b^2x^2 + a^2y^2 = a^2b^2 in the first quadrant and forms with the coodiante axes the triangle with smallest possible area (a and b are positive constants)

    im not sure how to do this problem, the teacher explained it but it was very confusing.
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  2. #2
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    Quote Originally Posted by viet View Post
    Find the equation of the line that is tangent to the ellipse
    b^2x^2 + a^2y^2 = a^2b^2 in the first quadrant and forms with the coodiante axes the triangle with smallest possible area (a and b are positive constants)

    im not sure how to do this problem, the teacher explained it but it was very confusing.
    Hello, viet,

    have a look here: http://www.mathhelpforum.com/math-he...html#post26090

    EB
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  3. #3
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    Hello, viet!

    Find the equation of the tangent to the ellipse [1] b^2x^2 + a^2y^2 \:= \:a^2b^2 in quadrant 1
    which forms with the coodiante axes the triangle with smallest possible area.
    ( a and b are positive constants.)

    Here is a little-known (but very convenient) formula . . .

    The equation of the tangent to the ellipse: . \frac{x^2}{a^2} + \frac{y^2}{b^2}\;=\;1 at the point (h,k)
    . . is given by: . \frac{hx}{a^2} +\frac{ ky}{b^2}\:=\:1

    Then the intercepts of this tangent are: . \left(\frac{a^2}{h},\,0\right) and \left(0,\,\frac{b^2}{k}\right)

    The area of the triangle is: . A \:=\:\frac{1}{2}\left(\frac{a^2}{h}\right)\left(\f  rac{b^2}{k}\right) \;= \;\frac{a^2b^2}{2hk}

    For convenience, let x = h,\:y = k. . Then we have: . A \;= \;\frac{a^2b^2}{2}x^{-1}y^{-1}

    Differentiate: . A' \;= \;\frac{a^2b^2}{2}\bigg[\left(x^{-1}\right)\left(\text{-}y^{-2}\right)y' + \left(\text{-}x^{-2}\right)\left(y^{-1}\right)\bigg]

    Equate to zero: . -\frac{y'}{xy^2} - \frac{1}{x^2y} \;= \;0

    Multiply by -x^2y^2:\;\;xy' + y \;=\;0\quad\Rightarrow\quad y \:=\:-xy' [2]

    . . The ellipse is: .[1] b^2x^2 + a^2y^2\:=\:a^2b^2
    . . Differentiate implicitily: . 2b^2x + 2a^2yy' \:=\:0\quad\Rightarrow\quad y' \:=\:-\frac{b^2x}{a^2y} [3]

    Substitute [3] into [2]: . y \:=\:-x\left(-\frac{b^2x}{a^2y}\right) \quad\Rightarrow\quad a^2y^2\:=\:b^2x^2 [4]

    Substitute [4] into [1]: . b^2x^2 + b^2x^2\:=\:a^2b^2\quad\Rightarrow\quad 2x^2 - a^2\quad\Rightarrow\quad x = \frac{a}{\sqrt{2}}

    Substitute into [1] and get: . y\,=\,\frac{b}{\sqrt{2}}

    Hence, the intercepts of the tangent are:
    . . \left(\frac{a^2}{\frac{a}{\sqrt{2}}},\:0\right) = (\sqrt{2}a,\,0) and \left(0,\:\frac{b^2}{\frac{b}{\sqrt{2}}}\right) = (0,\,\sqrt{2}b)

    And I'll let you write the equation of that tangent line . . .

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  4. #4
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    i got it, thanks for your help
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