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Math Help - Derivatives and Perpendicular Lines

  1. #1
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    Derivatives and Perpendicular Lines

    Determine the coordinates of the point of intersection of the two perpendicular lines that intersect on the y-axis and are both tangent to the parabola given below:

    y = 9x^2

    When I sketch this out, I can see visually that this point of intersection cannot occur at a positive y-value. Other things that are quickly apparent:

    f'(x) = 18x
    The point of intersection must have an x-coordinate of 0.

    So, if I use the formula of a line, and plug in an x-coordinate of 0, I get:

    y = f(0) + f'(0)(x - 0)
    y = 0 + 0(x - 0)
    y = 0

    And I conclude that my point of intersection is (0,0). All fine and dandy, except that this isn't the right answer. Can anybody help?

    Thanks.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    The two tangents should be at right angles to each other, and at the same time tangent to the curve.

    Let the gradient of the first line be m_1

    That of the second will be m_2

    And m_1 \times m_2 = -1
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    Yeah, I thought of that (and probably should have listed it in my original post), but I still don't know how to integrate it into this solution. Here's some further thoughts:

    Let L_1 = the positively sloping line and L_2 be the negatively sloping line.

    Let L_1 pass through the points, (x_1, y_1) and (0, y_2) where y_2 < 0.

    Since y = 9x^2 is even, this means that L_2 passes through (-x_1, y_1) and (0, y_2).

    As you point out, the slopes are negative reciprocals of each other, by definition.

    Then, there are a couple of things that I think are true, but I don't know just how to prove them. First m_1 = 1 and m_2 = -1. I say this because, in order to be tangent to an even function, the only way that I can see two perpendicular lines meeting at the y-axis would be if they both had symmetrical x-coordinates (e.g. (5, 0) and (-5, 0)).

    If that is true, this final one might be the key to the whole thing. If the slope is 1 or -1, and the lines intersect at the y-axis, then doesn't y_2 have to equal -(y_1)?

    Am I on the right track? Making this to complicated? Any help is appreciated.

    Thanks again.
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  4. #4
    MHF Contributor Unknown008's Avatar
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    That's right! m1 and m2 equal to -1 and 1 (order unimportant).

    So that, the gradient at anypoint being 18x, you get:

    18x = 1

    x = \dfrac{1}{18}

    And:

    18x = -1

    x = -\dfrac{1}{18}

    Those are your two x coordinates. In your picture, the points where the tangents meet the line have the same y coordinate.

    At x = 1/18, find y and from the points and slope, get the equation of the lines (in fact, one only is necessary)

    Then, find where the line meets the y axis.

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  5. #5
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    Quote Originally Posted by joatmon View Post
    Yeah, I thought of that (and probably should have listed it in my original post), but I still don't know how to integrate it into this solution. Here's some further thoughts:

    Let L_1 = the positively sloping line and L_2 be the negatively sloping line.

    Let L_1 pass through the points, (x_1, y_1) and (0, y_2) where y_2 < 0.

    Since y = 9x^2 is even, this means that L_2 passes through (-x_1, y_1) and (0, y_2).

    As you point out, the slopes are negative reciprocals of each other, by definition.

    Then, there are a couple of things that I think are true, but I don't know just how to prove them. First m_1 = 1 and m_2 = -1. I say this because, in order to be tangent to an even function, the only way that I can see two perpendicular lines meeting at the y-axis would be if they both had symmetrical x-coordinates (e.g. (5, 0) and (-5, 0)).

    If that is true, this final one might be the key to the whole thing. If the slope is 1 or -1, and the lines intersect at the y-axis, then doesn't y_2 have to equal -(y_1)?

    Am I on the right track? Making this to complicated? Any help is appreciated.

    Thanks again.
    you are on the right track ... the use of symmetry is key to this problem.
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  6. #6
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    Got it. Thanks to both of you for your help.

    f(\frac{1}{18}) = 9(\frac{1}{18})^2 = \frac{1}{36}

    So the tangental points are (\frac{1}{18}, \frac{1}{36}) and (\frac{-1}{18}, \frac{1}{36})

    And that means that the lines intersect at (0, \frac{-1}{36}).

    Can't thank you enough for your help.
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