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    I've tried doing this question but I keep getting it wrong: Answer is :
    12y= 12x - 17

    - Find the equation of the normal to the curve y=3x^2 - 2x - 1 which is parallel to the line y=x-3 .

    ----------------------

    I start off by finding dy/dx , which is 6x-2 . I set it equal to 1 (gradient of the parallel) to find the x co-ordinate.

    6x -2 = 1
    6x=3
    x= 0.5
    Y= -5/4

    Plug 0.5 back in to find the gradient, it gives 0 :confused:

    -----------------------------
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    Remember you're trying to find the normal, not the tangent.

    The gradient of the normal is perpendicular to that of the tangent.
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    (Original post by chembob)
    Remember you're trying to find the normal, not the tangent.

    The gradient of the normal is perpendicular to that of the tangent.
    Don't I do that after I get the gradient of the tangent at x ? then reciprocate it ?
    So it'd still be 0 ?
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    You're not finding the gradient. You're finding out where it equals 1. This is when x=1/2 and y=-5/4.

    So at (1/2,-5/4) the curve has a tangent with a gradient of 1. The normal here has a gradient of -1.
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    (Original post by Get me off the £\?%!^@ computer)
    You're not finding the gradient. You're finding out where it equals 1. This is when x=1/2 and y=-5/4.

    So at (1/2,-5/4) the curve has a tangent with a gradient of 1. The normal here has a gradient of -1.

    Okay so (1/2,-5/4) are the points, -1 is the gradient.

    Putting it in the formula of straight line :

    y+5/4 = -1 (x - 1/2)

    Multiply through by 4 to give:
    4y+5 = -4x - 2

    4x+4y+7 =0

    ----------

    Answer is meant to be 12y= 12x - 17 :confused:
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    (Original post by Gelato)
    Okay so (1/2,-5/4) are the points, -1 is the gradient.

    Putting it in the formula of straight line :

    y+5/4 = -1 (x - 1/2)

    Multiply through by 4 to give:
    4y+5 = -4x - 2 <-----------------------------------should be plus

    4x+4y+7 =0

    ----------

    Answer is meant to be 12y= 12x - 17 <----------------------------------- this clearly has the wrong gradient.

    :confused:
    Both answers are wrong but yours is better.
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    Argh not sure

    That's the answer in the back of the book
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    (Original post by Gelato)
    Argh not sure

    That's the answer in the back of the book
    Your coordinates are wrong.

    The gradient of the normal = 1 so the gradient of the tangent (perpendicular) at the required point = -1
    giving

    dy/dx = 6x - 2 = -1

    so x= 1/6 and you can substitute into the original equation to find y
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    Thanks for the help , I see where I went wrong .

    I got x as 1/6 Y as -7/6 gradient as -1

    Putting them in the formula gives:

    y+7/6 = -1 ( x - 1/6 )

    Multiplying through by 6 on the left:
    6y + 7 = -6 (x - 1)
    -------------------------

    I end up with 6x+ 6y = 13
 
 
 
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