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Equations of tangents

  1. Jan 7, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the equations of the tangents of the equation [tex] x^2+(y-4)^2=4[/tex] that pass through the origin.

    3. The attempt at a solution

    Ok, I don't know if im overcomplicating this (takes a deep breath):

    The equation of tangent that pass through the origin has the form
    And the derivative of the curve is given by
    [tex] \frac{dx}{dy}=\frac{1}{2}(4-(y-4)^2)^\frac{-1}{2}(2y+8)[/tex]
    Then equate m=dy/dx and y=mx into the equation of the curve

    Here is where it gets really confusing and where I start to doubt my workings.
  2. jcsd
  3. Jan 7, 2008 #2


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    Well, first of all though not strictly necessary, do note that the graph of the equation is actually a circle of radius 2 centred at (0,4). Once you get that, you should be able to visualise intuitively that because the circle is centred on y-axis, there should be only 2 tangent lines to the circle that passes through the origin.

    I don't think your working for dy/dx is correct. Differentiate the equation implicitly, don't bother expressing everything in terms of one variable x or y only. Then you'll be able to do the question easily.
  4. Jan 7, 2008 #3
    ok so i get

    [tex] 2x+2\frac{dy}{dx}-8\frac{dy}{dx}=0[/tex]

    and then the curve becomes

    [tex] x^2+\frac{1}{9}x^2-\frac{8}{3}x+12=0[/tex]

    so i solve for x which gives me the intercepts and i can solve for m is this correct?
  5. Jan 7, 2008 #4
    Actually impressively I believe his equation for dy/dx is almost right, or I screwed it up(possible!)

    if you do it implicity, you get 2x+2(y-4)dy/dx=0? so dy/dx=-2x/(2y-8)

    Unless I interpret this wrong though, that means the slope of the tangent lines at the origin is 0, which is obviously not the case, so I think I suck
    Edit: You slipped your post in before mine

    I believe the error you made is that the derivative of y(x)^2 is 2y*dy/dx, not just 2*dy/dx. With that in mind I think you get the same thing I did
    Last edited: Jan 7, 2008
  6. Jan 7, 2008 #5


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    Um no. Differentiating implicitly yields
    [tex]2x + 2(y-4) \frac{dy}{dx} = 0[/tex]
    Treat y as a function of x, and remember you are differentiating implicitly with respect to x.

    EDIT: Latex doesn't seem to be displaying the equation correctly. Blochwave's answer is correct.
    Last edited: Jan 7, 2008
  7. Jan 7, 2008 #6
    ugh scrap that careless mistakes

    so differentiating implicity yields


    hey i think i get back the same derivative as my OP
  8. Jan 7, 2008 #7


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    Where did you get [tex]2y\frac{dy}{dx} - 8\frac{dy}{dx}[/tex] from?

    Implicit differentiation should give [tex]2x + 2(y-4) \frac{dy}{dx} = 0[/tex] instead. You're using the chain rule here.
  9. Jan 7, 2008 #8
    He's right, he just skipped the chain rule and multiplied out that binomial squared

    if you factor out the dy/dx you get yours, and mine, answer
  10. Jan 7, 2008 #9


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    Oh yeah. Sheesh. Stupid mistake.
  11. Jan 7, 2008 #10
    ok so i substitute the derivative and I have this ugly monster


    I have no idea how to simplfiy this, which was why I thought my answer was wrong
  12. Jan 8, 2008 #11


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    How did you get that equation? You substituted the derivative dy/dx into where?
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