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Math Help - Normal line, parallel lines

  1. #1
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    Normal line, parallel lines

    find an equation of the normal line to y=x2-5x+4 that is parallel to the line x-3y=5.

    please help me with this!!
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Normal line, parallel lines

    One way to work this problem:

    Implicitly differentiate the parabola with respect to y and solve for -\frac{dx}{dy}. Equate this to the slope of the given line, allowing you to find the point of intersection of the normal line and the parabola. Then use the point-slope formula to find the equation of the normal line.
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    Re: Normal line, parallel lines

    What is the slope of the line x- 3y= 5? Since you want a line parallel to that, you want a line with that same slope.

    What is the derivative of x^2- 5x+ 4 for general x? That will be the slope, m, of the tangent line to the curve. A normal line will have slope -1/m. For what x is the derivative equal to -1 over the slope from the first question?
    Last edited by HallsofIvy; September 26th 2012 at 07:17 AM.
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    Re: Normal line, parallel lines

    the derivative of the the first equation is y'=2x-5 and the derivative of the second equation is y'=1/3 so now what do i do?
    Quote Originally Posted by HallsofIvy View Post
    What is the slope of the line x- 3y= 5? Since you want a line parallel to that, you want a line with that same slope.

    What is the derivative of x^2- 5x+ 4 for general x? That will be the slope, m, of the tangent line to the curve. A normal line will have slope -1/m. For what x is the derivative equal to -1 over the slope from the first question?
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    Re: Normal line, parallel lines

    You don't need the derivative of the second line in this problem. First put the second line in y = mx + b form. It has a slope of  1/3 , a line normal to that function would have a slope that's the negative reciprocal  -3 . The negative reciprocal or -1/m of 1/3 is -3. To answer HallofIvy's question,
    For what x is the derivative equal to -1 over the slope from the first question?
    we can equate the negative reciprocal of the second line  -3 with the derivative of the first function (which is what MarkFL2 was saying).  -3 = 2x -5 , this gives  x = 1 . So, if you plug 1 into the derivative of the parabola you'll get  -3 which is the negative reciprocal of the second line's slope. Plug 1 into the equation of the parabola and we'll get y = 0, so the point where the normal line intersects the parabola is (1, 0), and using the point slope form with the exact same slope of the second line  (y-0) = 1/3(x-1) gives  y = 1/3x -1/3 .

    I was working on a similar problem and I checked this site out here for help: Find the equation of the normal line of the given parabola that is parallel to a line
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