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Finding parametric equation of the line tangent to the parabola

  1. Sep 17, 2011 #1
    1. The problem statement, all variables and given/known data
    Find a parametric equation of the line that satisfies the condition:
    The line that is tangent to the parabola y=x^2 at the point (-2,4)

    3. The attempt at a solution
    My answer came out to
    <x,y> = <-2,4> + t<1,2>
     
  2. jcsd
  3. Sep 17, 2011 #2
    Any curve whose equation is given by y = f(x) can be parametrized as
    r = <x, f(x)>.
     
    Last edited: Sep 17, 2011
  4. Sep 17, 2011 #3

    dynamicsolo

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    Homework Helper

    For a parametric curve, the slope at a point (x0, y0) is

    [tex](\frac{dy}{dx}) \vert_{(x_{0},y_{0})} = \frac{dy/dt}{dx/dt} \vert_{(x_{0},y_{0})} .[/tex]

    What is dy/dx for the parabola at (-2, 4)? What is the slope of your line? (There is a discrepancy.) How did you work out your answer?


    glebovg, shemer77 has written the pair of equations as a single vector equation -- that notation is acceptable for a system of equations.
     
  5. Sep 17, 2011 #4
    sorry, im not really sure what your asking? What I did was I took the derivative of y=x^2 b/c that would give me the slope of a line tangent to the parobala then i used that slope as my vector. Are you asking whats the derivative at (-2,4)?
     
  6. Sep 17, 2011 #5

    HallsofIvy

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    You have told us, twice, what you got without telling us how you got it. The answer you got is NOT correct but since you haven't shown how you got that, we can't tell what you did wrong.
     
  7. Sep 17, 2011 #6
    I mentioned it in my last post

    Are you saying I should take the derivative which would give me dy/dx=2x and thus dy/dx=-4? and would I put that into my original equation? <x,y> = <-2,4> + t<1,-4>
     
  8. Sep 18, 2011 #7
    My answer looked completely different from yours and based on what HallsofIvy says, I think I am not wrong.

    What I did was do the line the normal way, that is

    Find y = y'(x - x_0) + y_0

    Then let x = t
     
  9. Sep 18, 2011 #8

    HallsofIvy

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    Just saying you "took the derivative" does not tell us how you got that equation! The "algebraic" form of the equation would be y= m(x-1)- 4 where m is the value of the derivative at (1, -4). One possible parametric equation for that would be t<1, m>+ <1, -4>. Do you see why?
     
  10. Sep 18, 2011 #9
    Hmm, I think so but I dont see how my answer of
    <x,y> = <-2,4> + t<1,-4>
    is wrong?
     
  11. Sep 18, 2011 #10
    That's odd I got q(t) = <0,-4> + t<1,m>
     
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