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Isosceles hyperbola locus problem

  1. Nov 6, 2012 #1
    Hello all,

    I would like beforehand to inform you that the translation of the following geometric problem is not very good and consequently you will have to use your mathematical intuition just a little bit. I encountered it while giving admission exams in a Mathematics department to pursue a second degree.
    I don't have a mathematical background so any help appreciated.

    1. The problem statement, all variables and given/known data
    "Find the locus of points from which the conducted tangents at a given isosceles hyperbola are vertical to each other"

    2. Relevant equations



    3. The attempt at a solution
    Can you assume that the locus contains just one point (which is the origin of the axes) and the tangents touch the curve at infinity ? :grumpy:
    I apologize for my English ...
     
    Last edited: Nov 6, 2012
  2. jcsd
  3. Nov 6, 2012 #2

    haruspex

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    All makes sense to me except for "isosceles hyperbola". Could it mean rectangular?
    For the rest, I would word it as "Find the locus of points from which the subtended tangents to a given (?) hyperbola are perpendicular to each other"
    No.
     
  4. Nov 7, 2012 #3
    Yes, I do mean rectangular hyperbola.
    When it comes to geometry it is all Greek to me ... :confused:

    I'm not sure whether we should say "conducted tangent from a given point" or "subtended tangent".

    As far as I understand two tangent lines start from a point on the locus and touch the given curve in a way that they are perpendicular to each other.

    Any suggestions ?
     
  5. Nov 7, 2012 #4

    haruspex

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    Let (x,y) be a point on the locus, (x',y'), (x", y") be the contact points on the hyperbola of the conducted/subtended tangents. We have the following equations (assuming a canonical form for the hyperbola):
    y'2-x'2=c2
    y"2-x"2=c2
    y'y - x'x = c2
    y"y - x"x = c2
    x'x"+y'y" = 0 (tangents perpendicular)
    In principle, should be possible to eliminate x', x", y', y" to obtain an equation for x and y.
     
  6. Nov 9, 2012 #5
    I see your point but there is always a but ... :frown:

    No matter how hard I try, I cannot figure out an equation by eliminating x',x'',y' and y''.

    Maybe there is another way of solving the problem.

    I don't know. What do you think ?
     
  7. Nov 9, 2012 #6

    haruspex

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    I worked it through and got x=y=0. Then I thought about it geometrically and realised that is right. The gradient of the curve y2=x2+c2 has magnitude <= 1 everywhere. So two tangents at right angles must have gradients +1 and -1. In the more general hyperbola y2=m2x2+c2, only if m>1 does the locus become more interesting.
     
  8. Nov 10, 2012 #7
    So, the answer is that there is no locus or the locus contains just one point and the contact points with the curve are at infinity as I initially proposed (?).

    If it is so, I must play lottery. :rolleyes:

    Thank you for your time.
     
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