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Math Help - Finding points on a parabola (involving normal lines)

  1. #1
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    Finding points on a parabola (involving normal lines)

    "Find the point(s) on the parabola f(x) = x^2 - 4x + 3 at which the normal line passes through the point (2,0). Illustrate by graphing the parabola and the normal line(s)."

    I started off by finding f'(x) = 2x - 4 which would be the slope of a tangent line at x.

    Next, I assumed the negative reciprocal (-1)/(2x - 4) is the slope of the normal line.

    Then, using the given point:

    y - 0 = ((-1)/(2x - 4))(x - 2)

    y = (-0.5)

    x^2 - 4x + 3 = (-0.5)

    x^2 - 4x + 3.5


    But when I use the quadratic formula I get:

    (4 + sqrt(2))/2 or (4 - sqrt(2))/2

    which appears to be a problem. I must have messed up somewhere...

    Any help with solving this problem is appreciated.
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  2. #2
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    Quote Originally Posted by Tulki View Post
    "Find the point(s) on the parabola f(x) = x^2 - 4x + 3 at which the normal line passes through the point (2,0). Illustrate by graphing the parabola and the normal line(s)."

    I started off by finding f'(x) = 2x - 4 which would be the slope of a tangent line at x.

    Next, I assumed the negative reciprocal (-1)/(2x - 4) is the slope of the normal line.


    *** Here's the problem: the tangent line's slope is simply 2, so its reciprocal is only -1/2 and not all the mess you wrote!
    Remember: slope is a number.

    Tonio



    Then, using the given point:

    y - 0 = ((-1)/(2x - 4))(x - 2)

    y = (-0.5)

    x^2 - 4x + 3 = (-0.5)

    x^2 - 4x + 3.5


    But when I use the quadratic formula I get:

    (4 + sqrt(2))/2 or (4 - sqrt(2))/2

    which appears to be a problem. I must have messed up somewhere...

    Any help with solving this problem is appreciated.
    ...
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  3. #3
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    I'm probably missing something obvious here, but how do you know the slope of the tangent line is 2?
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  4. #4
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    Sorry about the triple post, but now I'm not sure of this again.

    Using the normal line slope I get the normal line equation:

    y = (-1/2)x + 1

    but I can't seem to solve the rest of the question. Help!
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  5. #5
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    Quote Originally Posted by Tulki View Post
    Sorry about the triple post, but now I'm not sure of this again.

    Using the normal line slope I get the normal line equation:

    y = (-1/2)x + 1

    but I can't seem to solve the rest of the question. Help!
    Well, now you only need to find out where the parabola intersects the normal line y = (-1/2)x + 1. I didn't check it to depth, but I'm afraid you get some rather nasty answer here. Anyway, the points of intersection of the normal line and the parabola are obviously the points on the parabola that you're looking for.

    Tonio
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  6. #6
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    Ah. Thank you for the clarification. It's rare for us to receive problems with odd answers that require a calculator to compute, since they discourage it (we can't use them on tests, after all).

    Guess I'll grab the answers now!
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  7. #7
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    The slope of the normal line is not 2, you had it right.

    because when you do the derivative of a function it gives you f' which is the slope of a tangent line at any given point on the curve. So in this question when you do the derivative, it simply gives you a formula for the slope of the tangent at any point on the graph; You had the right idea.
    as far as I can see, the problem came in when you did your normal line.
    the equation for the normal line should be *after some manipulation*
    y=-(x-2)/(2x-4) which is -(x-2)/(x-2) and turns out to be simply -1.
    You then equate -1 to the formula for your parabola
    -1=x^2-4x+3 and solve for x.
    I'm doing the same question and getting stuck here, because after doign quadratic, it says my X should be zero, but for some reason it doesn't seem like it'd be the answer lol.
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