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Math Help - minding the ps and qs

  1. #1
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    Cool minding the ps and qs

    Hi guys!

    I am baffled by my own solution to the following parabola question. Having got the right answer, I find that I don't like it! Can anyone help?

    The parabola is y^2 = 4ax and at point P(ap^2, 2ap) the tangent is found to be yp = x + ap^2. We have to find the coordinates of the point of intersection T of the tangent at P and the tangent at Q(ap^2, 2aq).

    In the derivation of the tangent at P the gradient is found to be 1/p.
    To get the tangent at Q, given that P and Q have the same coordinate ap^2, I would expect the intersection to be on the x-axis (since y^2 = 4ax is symmetrical). The answer does not confirm this, so that's my first question - why not?

    If I put the coords of Q into the parabola I get:

    (2aq)^2 = 4a(ap^2)
    q^2 = p^2
    q = \pm p

    I therefore take the gradient at Q to be -1/p, then the equation of the tangent follows:

    \frac{y - 2aq)}{x - ap^2} = \frac{-1}{p}
    call this equation 1
    yp - 2apq = -x + ap^2
    The tangent at P is
    yp = x + ap^2
    subtracting gives
    x = apq
    y = a(p + q)
    which is the correct answer.

    The thing that puzzles me is that to get the answer I am assuming the gradient at Q is symmetrical i.e. -1/p (since the gradient at P is 1/p). Also if I express this gradient as 1/q which is the same thing and substitute into equation 1, I get:

    \frac{y - 2aq)}{x - ap^2} = \frac{1}{q}
    yq - 2aq^2 = x - ap^2
    and equation 1 is:
    yp = x + ap^2
    subtracting:
    y(p - q) - 2aq^2 = -2ap^2
    y(p - q) = 2a(q + p)(q - p)
    y = 2a(q + p)

    A different result! All these ps and qs are swimming before my eyes. Have I made a slip somewhere?
    Last edited by s_ingram; July 28th 2009 at 03:16 AM. Reason: Latex syntax error
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  2. #2
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    Quote Originally Posted by s_ingram View Post
    Hi guys!

    I am baffled by my own solution to the following parabola question. Having got the right answer, I find that I don't like it! Can anyone help?

    The parabola is y^2 = 4ax and at point P(ap^2, 2ap) the tangent is found to be yp = x + ap^2. We have to find the coordinates of the point of intersection T of the tangent at P and the tangent at Q(ap^2, 2aq).
    I presume that q is not p but then q= -p. So Q is the point (ap^2, -2ap).

    In the derivation of the tangent at P the gradient is found to be 1/p.
    To get the tangent at Q, given that P and Q have the same coordinate ap^2, I would expect the intersection to be on the x-axis (since y^2 = 4ax is symmetrical). The answer does not confirm this, so that's my first question - why not?
    If I am understanding what you wrote correctly, yes, the point of intersection of the tangents at P(ap^2, 2ap) and Q(ap^2, -2ap) intersect on the x-axis.

    If I put the coords of Q into the parabola I get:

    (2aq)^2 = 4a(ap^2)
    q^2 = p^2
    q = \pm p

    I therefore take the gradient at Q to be -1/p, then the equation of the tangent follows:

    \frac{y - 2aq)}{x - ap^2} = \frac{-1}{p}
    call this equation 1
    yp - 2apq = -x + ap^2
    Okay. But q= -p so this is the same as yp- 2ap^2= -x+ ap^2

    The tangent at P is
    yp = x + ap^2
    subtracting gives
    x = apq
    y = a(p + q)
    which is the correct answer.
    And, of course, q= -p so this is x= ap^2 and y= 0

    The thing that puzzles me is that to get the answer I am assuming the gradient at Q is symmetrical i.e. -1/p (since the gradient at P is 1/p). Also if I express this gradient as 1/q which is the same thing and substitute into equation 1, I get:

    \frac{y - 2aq)}{x - ap^2} = \frac{1}{q}
    yq - 2aq^2 = x - ap^2
    and equation 1 is:
    yp = x + ap^2
    subtracting:
    y(p - q) - 2aq^2 = -2ap^2
    y(p - q) = 2a(q + p)(q - p)
    y = 2a(q + p)

    A different result! All these ps and qs are swimming before my eyes. Have I made a slip somewhere?
    No, it's not a different result. q is still -p, q+ p= 0, and 2a(0)= a(0)= 0. The only "slip" you made was not to replace q by -p as soon as you realized, as you say above, that q= \pm p. Of course, q= p does not give two points so we don't have two tangent to intersect and the whole question falls apart: q= -p.
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    You have unconfused me! Thank you.
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    Quote Originally Posted by s_ingram View Post
    Hi guys!

    I am baffled by my own solution to the following parabola question. Having got the right answer, I find that I don't like it! Can anyone help?

    The parabola is y^2 = 4ax and at point P(ap^2, 2ap) the tangent is found to be yp = x + ap^2. We have to find the coordinates of the point of intersection T of the tangent at P and the tangent at Q(ap^2, 2aq).
    Are you sure that you have copied the question correctly? It seems more likely to me that the point Q should be (a{\color{red}q}^2, 2aq). Then the tangent at P has equation yp = x + ap^2 and the tangent at Q has equation yq = x + aq^2. Subtract the second of those equations from the first, divide by pq, and you get the answer y=a(p+q).
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    Quote Originally Posted by Opalg View Post
    Are you sure that you have copied the question correctly? It seems more likely to me that the point Q should be (a{\color{red}q}^2, 2aq). Then the tangent at P has equation yp = x + ap^2 and the tangent at Q has equation yq = x + aq^2. Subtract the second of those equations from the first, divide by pq, and you get the answer y=a(p+q).
    No, the problem is clearly for the two points where a vertical line crosses the parabola. Otherwise the whole problem makes no sense.
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    No, the problem is clearly for the two points where a vertical line crosses the parabola. Otherwise the whole problem makes no sense.
    On the contrary, the problem makes perfect sense if P=(ap^2,2ap) and Q=(aq^2,2aq). The tangents at these two general points on the parabola meet at the point (apq,a(p+q)), whose y-coordinate is a(p+q), as required.

    If the problem refers to two points where a vertical line crosses the parabola, then we would have to have q=p, and the tangents would intersect on the x-axis. But the answer indicates that this is not the situation, as the OP noted:
    Quote Originally Posted by s_ingram View Post
    To get the tangent at Q, given that P and Q have the same coordinate ap^2, I would expect the intersection to be on the x-axis (since y^2 = 4ax is symmetrical). The answer does not confirm this, so that's my first question - why not?
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  7. #7
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    S_ingram didn't say what "answer" he was referring to. I thought he meant the answer he had got to the problem: y= p+ q. Of course, your interpretation is reasonable- assuming that he had copied the problem incorrectly.
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