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Math Help - Intersection of Line and Hyperbola

  1. #1
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    Intersection of Line and Hyperbola

    A hyperbola is intersected by line 1 at (u_1, v_1) and (v_1, u_1)
    and by line 2 at (u_2, v_2) and (-v_2, -u_2)
    Lines 1 and 2 intersect at (u_1+v_1, 0)

    These can be written

    u_1+v_1=u_2-v_2\ \ \ \ \ \ \(1)
    u_1v_1=u_2v_2\ \ \ \ \ \ \ \ \ \ \ \ \ (2)

    I am trying to find a general solution for the values of u_1, v_1, u_2 and v_2.

    I have tried squaring (1) and adding\subtracting various multiples of (2), but this just brings me various fractions of the starting values.

    Any help or pointers would be greatly appreciated
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  2. #2
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    re: Intersection of Line and Hyperbola

    ok,

    Here's where I am so far...

    u_1+v_1=u_2-v_2

    u_1v_1=u_2v_2

    let\ \ \ u_2=u_1+x

    and\ \ v_2=v_1-y

    then we need

    (u_1+x)(v_1-y)=u_1v_1

    v_1x-xy-u_1y=0

    y=\dfrac{v_1x}{u_1+x}

    x=\dfrac{u_1y}{v_1-y}

    Is this as far as I can go?
    Last edited by moriman; September 29th 2011 at 05:45 AM.
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  3. #3
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    Re: Intersection of Line and Hyperbola

    I had forgotten about one of the original constraints

    u_1+v_1=u_2-v_2

    This works out to x=2v_1-y

    Taking this with the x from the previous result I have

    2v_1-y=\dfrac{u_1y}{v_1-y}

    This gives

    y=\dfrac{(3v_1+u_1)\pm\sqrt{u_1^2+6u_1v_1+v_1^2}}{  2}

    So for integer solutions (u_1^2+6u_1v_1+v_1^2) must be a perfect square.
    Last edited by moriman; September 29th 2011 at 12:21 PM. Reason: correcting error
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Intersection of Line and Hyperbola

    I have red your posts and it's confusing in my opinion, what do you want to calculate or show? ...
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  5. #5
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    Re: Intersection of Line and Hyperbola

    Hi,

    With the original equations

    u_1+v_1=u_2-v_2

    u_1v_1=u_2v_2

    The only way to find values for u_1, v_1, u_2, v_2 is by trial and error.

    I am trying to find a general solution.

    It is similar to the diophantine solution for a pythagorean triple
    In the attempt I've made so far, the

    y=\dfrac{(3v_1+u_1)\pm\sqrt{u_1^2+6u_1v_1+v_1^2}}{  2}

    and whatever x will work out to be ( a similar expression to y)
    If I can make sure that x and y are integers, I would have correct numbers to satisfy the original equations
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  6. #6
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    Re: Intersection of Line and Hyperbola

    OK, I'll try giving an example of what I'm looking for

    Take a^2+b^2=c^2

    If one sets

    a=m^2-1 and b=2m then c=m^2+1

    By choosing any integer m>1 a, b and c will be a pythagorean triple.
    This doesn't give all pythaorean triples, but it does give an infinite amount of them, which is usually enough

    I'm looking for a similar way of getting numbers that satisfy the 2 equations in the op.

    u_1+v_1=u_2-v_2

    u_1v_1=u_2v_2

    I had made further progress from my last post after I noticed that

    u_2 is not only greater than u_1, but also u_2>(u_1+v_1)

    so I set\ \ \ \ \ u_2=(u_1+v_1)+x

    and\ \ \ \ \ \ \ \ \ \ v_2=x

    This satisfies the first equation. When solving with the second equation I would have

    u_1+v_1=\dfrac{m-n}{2}

    u_1v_1=\dfrac{mn}{4}

    and\ \ \ x=\dfrac{n}{2}

    which is no further forward

    Quick synopsis of where I am at the moment

    u_1+v_1=u_2-v_2

    u_1v_1=u_2v_2

    u_2-(u_1+v_1)=\dfrac{u_1v_1}{u_2}

    u_2^2-(u_1+v_1)u_2-u_1v_1=0

    u_2=\dfrac{(u_1+v_1)\pm\sqrt{(u_1+v_1)^2+4u_1v_1}}  {2}

    similarly

    v_2=\dfrac{-(u_1+v_1)\pm\sqrt{(u_1+v_1)^2+4u_1v_1}}{2}

    Solving for (u_1+v_1)^2+4u_1v_1 being a perfect square leads me back to the beginning.
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