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Math Help - Proof involving hyberbola

  1. #16
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    Re: Proof involving hyberbola

    Yes you ar on the right track but to get from line 1 to line 2 you were multiplying throughout by a^2b^2. So instead of =1 you should have =a^2b^2
    ax^2+bx+c=0 has repeated roots if b^2-4ac=0 so apply this to your corrected equation.
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  2. #17
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    Re: Proof involving hyberbola

    You are multiplying both sides of the equation by a^2b^2 so in your second line you should have =a^2b^2 instead of =1
    The quadratic ax^2+bx+c=0 has repeated roots if b^2-4ac=0 so apply this to your quadratic.
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  3. #18
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    Re: Proof involving hyberbola

    OK, here goes:

    a = (b2 - m2a2)
    b = 2mca2
    c = (a2b2 - c2)

    For repeated roots, b2 = 4ac

    4m2c2a2 = 4.(b2 - m2a2)(a2b2 - c2)

    If I try to simplify and solve this equation, I don't get the result b2 - 4ac = 0

    If I simply ignore the constants a,b & c, I get the equation x2+2x+1 = 0, for which I get the roots (x+1) & (x+1)

    Is this correct?
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  4. #19
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    Re: Proof involving hyberbola

    yes right track but your second line should be=a^2b^2 (You were multiplying both sides by a^2b^2)
    Now the general quadratic ax^2+bx+c=0 has repeated roots if b^2-4ac=0
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  5. #20
    Don
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    Re: Proof involving hyberbola

    So,

    x^2(b^2 - m^2a^2) + x(-2mca^2) + (-a^2c^2- a^2b^2) = 0

    a = (b2 - m2a2)
    b = -2mca2
    c = (- a2c2 - a2b2)

    b2 = 4ac

    4m2c2a4 = 4.(b2 - m2a2).(-a2c2 - a2b2)

    4m2c2a4 = 4[ -a2b2c2 + m2a4c2 - a2b4 + m2a4b2]

    The two sides are not equal.

    We get 4m2c2a4 on both sides, but what to do with the other terms?
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  6. #21
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    Re: Proof involving hyberbola

    That 1 in your second line should be a^2b^2 (You were in the process of multiplying both sides by a^2b^2
    So line should read (b^2-a^2m^2)x^2-2a^2mcx-a^2(b^2+c^2)=0
    Now ax^2+bx+c=0 has repeated roots if b^2-4ac=0
    So in our equation want (2a^2mc)^2+4(b^2-a^2m^2)a^2(b^2+c^2)=0
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  7. #22
    Don
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    Re: Proof involving hyberbola

    That 1 in your second line should be a^2b^2 (You were in the process of multiplying both sides by a^2b^2
    So line should read (b^2-a^2m^2)x^2-2a^2mcx-a^2(b^2+c^2)=0
    Now ax^2+bx+c=0 has repeated roots if b^2-4ac=0
    So in our equation want (2a^2mc)^2+4(b^2-a^2m^2)a^2(b^2+c^2)=0

    I have carried out the correction of multiplying RHS also by a^2b^2 and started with the corrected equation,

    x^2(b^2 - m^2a^2) + x(-2mca^2) + (-a^2c^2- a^2b^2) = 0

    which can be rewritten as

    x^2(b^2 - m^2a^2) + x(-2mca^2) -a^2(c^2 + b^2) = 0

    I have carried out the multiplication as you have said. Incidentally, the equation should be

    (2a^2mc)^2-4(b^2-a^2m^2)a^2(b^2+c^2)=0

    and not

    (2a^2mc)^2+4(b^2-a^2m^2)a^2(b^2+c^2)=0

    because the term "C" is negative.

    which I have correctly done. How does all this relate to the original equation p^2 = a^2cos^2 A - b^2 sin ^2 A?

    Edit: Whether the term is negative or positive, it does not matter since there are uncancelled terms on the RHS.
    Last edited by Don; April 5th 2012 at 11:41 AM.
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  8. #23
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    Re: Proof involving hyberbola

    This is correct. Now multiply out brackets. There will be an a^2 which can be cancelled out of every term (since right hand side=0)
    You should also see among the terms a^2m^2c^2-a^2m^2c^2, so these go.
    Now you should see that there is b^2 which can be cancelled.
    Should now have b^2+c^2-a^2m^2=0
    Now replace c and m by what we let them stand for at the beginning.
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  9. #24
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    Re: Proof involving hyberbola

    OK, I got it, thanks again for your patience.
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  10. #25
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    Re: Proof involving hyberbola

    Well done! Incidentally in your last post you said that the 'C' is negative which is correct but the formula for repeated roots is B^2-4AC so we had two minus signs which is why I finished with a plus sign.
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