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Math Help - Finding point where normal line interesects parabola

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    Finding point where normal line interesects parabola

    The problem is: at what point does the normal to at intersect the parabola a second time?

     y ' = 8x +3

    Then can I just take the slope of 8, inverse it to be  -1/8 and then put it in point slope form,  (y-9)=-1/8(x-1)
    This ends up being  y = -1/8x + 73/8 .

    Then do I make it  -1/8x + 73/8 = 4x^2 + 3x + 2 ? The math gets pretty hairy after that, and I just would like to know if I'm going in the right direction!
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    Re: Finding point where normal line interesects parabola

    Quote Originally Posted by AZach View Post
    The problem is: at what point does the normal to at intersect the parabola a second time?

     y ' = 8x +3

    Then can I just take the slope of 8, inverse it to be  -1/8 and then put it in point slope form,  (y-9)=-1/8(x-1)
    This ends up being  y = -1/8x + 73/8 .

    Then do I make it  -1/8x + 73/8 = 4x^2 + 3x + 2 ? The math gets pretty hairy after that, and I just would like to know if I'm going in the right direction!
    No you CAN'T just take the slope as being 8. The derivative gives you a FUNCTION which gives you the gradient of your curve at ANY point. It CHANGES as you change your point. The gradient at x = 1 is 8(1) + 3 = 11.
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    Re: Finding point where normal line interesects parabola

    Ah, that definitely explains my first problem.

    Then the tangent line is  y = 11x + 9 and the normal line is  (y-9) = -1/11(x-1) or  y = -1/11x + 100/11

    Next step, I substitute  -1/11x + 100/11 = 4x^2 + 3x + 2 since one of the equations is already equal to y.

    Multiply across by 11 to clear fractions  -x + 100 = 44x^2 + 33x + 22 ->  0 = 44x^2 + 34x - 78 -> Divide the equation by 2  0 = 22x^2 + 17x -39 .

    My next question, is the quadratic formula the next way to go? I've already got it computed, but I'm not sure what the answer means or if it's correct. I know the objective is to find the perpendicular line to the parabola and find the second intersection point. So, the quadratic formula should give me the x coordinates for the intersection points on the parabola, how do I get the f(x) coordinates?
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    Re: Finding point where normal line interesects parabola

    Okay, so I get it now. I use the quad. formula and got x = 1 (makes sense, this is the point given in the question and corresponds with a output of 9), and x = -1.772727273. I plugged the latter into the original function of the parabola and got 9.252066119, so the normal line intersects the parabola a second time at (-1.772727273, 9.252066119). (1,9) is where it intersects the first time.

    Is this kind of information related to finding tension in mechanics? Like if there's a swing or pendulum, could finding the intersection points of the perpendicular tell where the machine is stressed or receiving any kind of instantaneous change?
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