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Thread: Tangent to hyperbola

  1. #1
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    Tangent to hyperbola

    Prove that the lines $\displaystyle y=mx\pm a\sqrt{m^2-1}$ are tangent to the hyperbola $\displaystyle x^2-y^2=a^2$ for all values of m.

    My work:
    I substituted $\displaystyle y=mx\pm a\sqrt{m^2-1}$ into the equation.
    $\displaystyle x^2-m^2x^2\mp 2am\sqrt{m^2-1}x-a^2(m^2-1)=a^2$
    $\displaystyle (1-m^2)x^2\mp 2am\sqrt{m^2-1}x-a^2(m^2-2)=0$
    If $\displaystyle y=mx\pm a\sqrt{m^2-1}$ is a tangent to the hyperbola,
    $\displaystyle b^2=4ac$
    $\displaystyle \rightarrow(2am\sqrt{m^2-1})^2=-4(1-m^2)(m^2-2)$
    I solved for m and got $\displaystyle \pm 1$
    So this is definitely wrong, but I don't know what is wrong.
    Thanks for any help!
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  2. #2
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    Check your b^2 = 4ac calculation.
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  3. #3
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    this is my working:
    $\displaystyle 4a^2m^2(m^2-1)=-4a^2(1-m^2)(m^2-2)$
    $\displaystyle 4a^2m^4-4a^2m^2=-4a^2(3m^2-2-m^4)$
    $\displaystyle 4a^2m^4-4a^2m^2=-12a^2m^2+8a^2+4a^2m^4$
    $\displaystyle 8a^2m^2=8a^2$
    $\displaystyle m^2=\frac{8a^2}{8a^2}=1$
    $\displaystyle m=\pm 1$
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  4. #4
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    The last term is -a^2*m^2 + a^2 not -a^2*m^2 -a^2
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  5. #5
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    huh? i went through all my workings again and could not see any mistakes which when corrected would give me that result.
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  6. #6
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    Quote Originally Posted by arze View Post
    huh? i went through all my workings again and could not see any mistakes which when corrected would give me that result.






    The above equation becomes
    x^2(1-m^2) + 2amsqrt(m^2-1) -a^2*m^2 = 0
    In this b^2 = 4ac.
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  7. #7
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    oh! ok i see where i got it wrong. thanks!
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