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Math Help - Tangent to an ellipse

  1. #1
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    Tangent to an ellipse

    Show by implicit differentiation that the tangent to the ellipse:

    x^(2) / a^(2) + y^(2) / b^(2) = 1

    at the point (x0, y0) is:

    [ (x0x) / a^(2) ] + [ y0y / b^(2) ]

    I got my gradient to be y' = -b^(2)x0 / a^(2)y0 but my tangent line was:

    y = [ -b^(2)x0 / a^(2)y0 ] + [b^(2)x0^(2) / a^(2)y0^(2)]

    Im not sure what the problem may be as I simply subbed in the gradient and (x0,y0) points into the y = mx + c equation
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  2. #2
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    While trying to answer the following I transcribe part of the questions into LaTeX for clearity's sake


    Quote Originally Posted by TsAmE View Post
    Show by implicit differentiation that the tangent to the ellipse:

    x^(2) / a^(2) + y^(2) / b^(2) = 1


    *** \frac{x^2}{a^2}+\frac{y^2}{b^2}=1


    at the point (x0, y0) is:

    [ (x0x) / a^(2) ] + [ y0y / b^(2) ]


    *** \frac{x_0x}{a^2}+\frac{y_0y}{b^2}=1 . Please do note the correction "=1" done to the above


    I got my gradient to be y' = -b^(2)x0 / a^(2)y0 but my tangent line was:

    y = [ -b^(2)x0 / a^(2)y0 ] + [b^(2)x0^(2) / a^(2)y0^(2)]

    *** y=\frac{-b^2x_0}{a^2y_0}+\frac{b^2x_0^2}{a^2y_0^2}

    We agree: y'=\frac{dy}{dx}=-\frac{b^2x}{a^2y} and at the point (x_0, y_0) we get what you wrote, but what you say is the tangent line's eq. cannot be

    correct since, for example, that has no x-term in it...! The eq. of the tangent line to the ellipse at this point is in fact:

    y-y_0=-\frac{b^2x_0}{a^2y_0}(x-x_0)\Longrightarrow a^2y_0y=-b^2x_0(x-x_0)+a^2y_0 \Longrightarrow a^2y_0y+b^2x_0x=b^2x_0^2+a^2y_0^2 , and now just

    divide both sides by a^2b^2 and remember that (x_0,y_0) is a point on the ellipse...and you'll get what you were asked!

    Tonio



    Im not sure what the problem may be as I simply subbed in the gradient and (x0,y0) points into the y = mx + c equation
    .
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  3. #3
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    So you cant work it out with the y = mx + c equation? I carried on from your solution and got (y0y / b^2) + (x0x / a^2) = (x0 / a^2) + (y0 / b^2) but the right hand side doesnt equal 1?
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  4. #4
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    Of course you can use "y= mx+ c". As Tonio said, y', at (x_0, y_0) is equal to -\frac{b^2x_0}{a^2y_0} so an equation with that slope passing through [tex](x_0, y_0) would be y= -\frac{b^2x_0}{a^2y_0}(x- x_0)+ y_0

    Multiply that out and you will get "y= mx+ c".
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  5. #5
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    Yeah but if I use y = mx + c and sub in the gradient and the x0 and y0 then the "x" and "y" essentially disappear, but the tangent line is suppose to contain an x and y
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  6. #6
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    Quote Originally Posted by TsAmE View Post
    Yeah but if I use y = mx + c and sub in the gradient and the x0 and y0 then the "x" and "y" essentially disappear, but the tangent line is suppose to contain an x and y

    Uuh?? What do you mean by "the tangent line is supposed to contain x and y"?? What is not clear in the post I sent you??

    Tonio
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  7. #7
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    Quote Originally Posted by TsAmE View Post
    Yeah but if I use y = mx + c and sub in the gradient and the x0 and y0 then the "x" and "y" essentially disappear, but the tangent line is suppose to contain an x and y
    I have no idea what you mean by this. "Sub in x0 and y0" where? Not for x and y- x0 and y0 should appear only in m and c.
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