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Math Help - Tangent line to parabola

  1. #1
    C.C
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    Tangent line to parabola

    y=1/2x^2
    given point= (4,8)

    I wasn't in my class when they did this but this is what I did.

    Line form of point = y=m(x-4)+8
    1/2x^2= m(x-4)+8
    0=m(x-4)-1/2x^2+8
    0=m(x-4)- x^2 +16
    0=m(x-4) - (x-4) (x+4)
    Here's where I get confused.. My friend told me at this point I do grouping or something and yeah... I got this
    0=(m + (x+4) (x-4)
    0=(m+(4+4)) (x-4)
    m=8
    then I plug the m into the equation and get y=8(x-4)+8 for the equation. Can anyone check for me if it's right or wrong?
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  2. #2
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    Quote Originally Posted by C.C View Post
    y=1/2x^2
    given point= (4,8)

    I wasn't in my class when they did this but this is what I did.

    Line form of point = y=m(x-4)+8
    1/2x^2= m(x-4)+8
    0=m(x-4)-1/2x^2+8
    0=m(x-4)- x^2 +16 ... you forgot to multiply m(x-4) by 2
    0=m(x-4) - (x-4) (x+4)
    Here's where I get confused.. My friend told me at this point I do grouping or something and yeah... I got this
    0=(m + (x+4) (x-4)
    0=(m+(4+4)) (x-4)
    m=8
    then I plug the m into the equation and get y=8(x-4)+8 for the equation. Can anyone check for me if it's right or wrong?
    \frac{1}{2}x^2 = m(x-4)+8

    x^2 = 2m(x-4) + 16

    x^2 - 16 - 2m(x-4) = 0

    (x+4)(x-4) - 2m(x - 4) = 0

    (x-4)[(x+4) - 2m] = 0

    x = 4 , m = 4

    y = 4(x - 4) + 8
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