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Math Help - Tangent to ellipse

  1. #1
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    Tangent to ellipse

    Can some one help in finding the two lines that are tangent to the ellipse

    2x2 4x + y2 + 1 = 0

    Thank you...
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  2. #2
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    There are infinitely many lines that are tangent to an ellipse, so which 2 do you want?

    Also it would be much easier to understand your problem and equation if you learn latex or at least use the proper notions of symbols like ^ means to the power of...

    For instance this:

    2x^2 -4x + y^2 + 1 = 0 or
    this
    2x^2 -4x + y^2 + 1 = 0
    would have been easier to understand

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  3. #3
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    Tangent to ellipse

    x2 and y2 should be read as to the power of 2. Further, the tangents lines pass through the origin.

    My apologies...
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by scarecrow13 View Post
    Can some one help in finding the two lines that are tangent to the ellipse

    2x2 – 4x + y2 + 1 = 0

    Thank you...
    Let (x_0,y_0) be a point on the ellipse.

    Thus, the derivative is 4x-4+2y\frac{\,dy}{\,dx}=0\implies \frac{\,dy}{\,dx}=\frac{2(1-x)}{y}

    Thus, \left.\frac{\,dy}{\,dx}\right|_{(x_0,y_0)}=\frac{2  (1-x_0)}{y_0}

    Thus, the tangent line has the form y-y_0=\left[\frac{2(1-x_0)}{y_0}\right](x-x_0)

    Since the lines pass through the origin, we see that {\color{red}-y_0^2}=-2(x_0-x_0^2)

    What do we do now?

    Recall that 2x^2-4x+y^2+1=0. Thus, at the point (x_0,y_0), we get 2x_0^2-4x_0+y_0^2+1=0\implies 2x_0^2-4x_0+1={\color{red}-y_0^2}

    Thus, substituting this into the mess we came up for the tangent lines, we see that 2x_0^2-4x_0+1=-2(x_0-x_0^2)\implies 1=2x_0\implies {\color{red}x_0=\tfrac{1}{2}}

    Substitution this into the expression for -y_0^2, we get that 2\left(\tfrac{1}{2}\right)^2-4\left(\tfrac{1}{2}\right)+1=-y_0^2\implies -\tfrac{1}{2}=-y_0^2\implies {\color{red}y_0=\pm\tfrac{\sqrt{2}}{2}}

    Thus, at the point \left(\tfrac{1}{2},-\tfrac{\sqrt{2}}{2}\right), the tangent line has the equation y+\tfrac{\sqrt{2}}{2}=-\sqrt{2}\left(x-\tfrac{1}{2}\right)\implies y=-\sqrt{2}x+\tfrac{\sqrt{2}}{2}-\tfrac{\sqrt{2}}{2}\implies\color{red}\boxed{y=-\sqrt{2}x}


    Thus, at the point \left(\tfrac{1}{2},\tfrac{\sqrt{2}}{2}\right), the tangent line has the equation y-\tfrac{\sqrt{2}}{2}=\sqrt{2}\left(x-\tfrac{1}{2}\right)\implies y=\sqrt{2}x-\tfrac{\sqrt{2}}{2}+\tfrac{\sqrt{2}}{2}\implies\co  lor{red}\boxed{y=\sqrt{2}x}

    Does this make sense?

    --Chris
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  5. #5
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    Question tangent line to the ellipse

    I need help with this one.
    Find the tangent line(s) to the ellipse x^2 + 2y^2 +4x = 5 at the point (1,0)

    I differentiated the equation implicitly to get this: (-2x-4)/4y
    And thats where i got stuck. cuz i get a ZERO in the denoinator.
    well,, using analytic math I got this equation (not sure if its correct either): x+5/3=0
    but i HAVE TO get there using differentiation

    the teacher said something like this is gonna appear on the test, so please help
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