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Thread: Another Coordinate system!

  1. #1
    Junior Member rednest's Avatar
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    Post Another Coordinate system!

    Question

    Prove that the line with equation $\displaystyle y = mx + c$ is a tangent to the ellipse with equation $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ if, and only if, $\displaystyle a^2m^2 + b^2 = c^2$.

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  2. #2
    Super Member wingless's Avatar
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    The line $\displaystyle y = m\cdot x + C$ is tangent to $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

    There are two conditions:
    1. The line and the ellipse must have a common point (x,y)
    2. Their slopes must be equal at (x,y)

    The slope of $\displaystyle y = m\cdot x + C$ is m.

    And the ellipses is,
    $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
    (Implicit differentiation)
    $\displaystyle \frac{2x}{a^2} + \frac{2yy'}{b^2} = 0$
    (Some algebra)
    $\displaystyle y'= -\frac{x b^2}{y a^2}$

    As the slopes are equal, we can write
    $\displaystyle y'=m= -\frac{x b^2}{y a^2}$

    $\displaystyle y= -\frac{x b^2}{m a^2}$

    Plug this is $\displaystyle y=m\cdot x + C$

    $\displaystyle -\frac{x b^2}{m a^2}=m\cdot x + C$

    $\displaystyle x= -\frac{m c a^2}{m^2 a^2 + b^2}$

    Then plug it in $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

    $\displaystyle \frac{x^2}{a^2} + \frac{x^2 b^2}{m^2 a^4} = 1$

    $\displaystyle x^2 = \frac{1}{\frac{1}{a^2}+\frac{b^2}{m^2 a^4}}$

    $\displaystyle x^2 = \frac{m^2 a^4}{m^2 a^2 + b^2}$

    Plug the x we found before,

    $\displaystyle {\left ( -\frac{m c a^2}{m^2 a^2 + b^2} \right )}^2 = \frac{m^2 a^4}{m^2 a^2 + b^2}$

    $\displaystyle \frac{\not{m^2} c^2 \not{a^4}}{(m^2 a^2 + b^2)^{\not 2}} = \frac{\not{m^2} \not{a^4}}{\not m^2 \not a^2 + \not b^2}$

    $\displaystyle \boxed{~c^2 = a^2 m^2 + b^2~}$
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