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Math Help - Proof involving hyberbola

  1. #1
    Don
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    Proof involving hyberbola

    Prove that the line touches the hyperbola if .

    How do I start?

    Edit: Sorry for the error in the question.
    Last edited by Don; March 31st 2012 at 07:17 PM.
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  2. #2
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    Re: Proof involving hyberbola

    Check the wording of the question. What you've written isn't a line.
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  3. #3
    Don
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    Re: Proof involving hyberbola

    Corrected the question.
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    Re: Proof involving hyberbola

    Again have you copied the question correctly. That is an ellipse not a hyperbola.
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  5. #5
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    Re: Proof involving hyberbola

    biffboy, thanks for your patience. Error again and corrected again.
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    Re: Proof involving hyberbola

    Make x the subject of the line equation and substitute this into the hyperbola equation. Rearrange this as a quadratic in y. For the line to be a tangent we require this quadratic to have repeated roots. Applying this condition should produce the result stated.
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  7. #7
    Don
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    Re: Proof involving hyberbola

    So, do I substitute the value of y in the line equation with the value of y with respect to x?

    In that case, the equation becomes,

    Proof involving hyberbola-codecogseqn.gif

    So the line equation will become

    Proof involving hyberbola-codecogseqn.gif

    I tried to simplify the equation and it is becoming messy. Am I doing things right?

    (Sorry for the small image, laTEX is pain and I do not know how to put the special characters here. Tried to use an online latex editor but there seems to be some problem)
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    Re: Proof involving hyberbola

    Replace y in the hyperbola equation with (p-xcosa)/sina
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  9. #9
    Don
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    Re: Proof involving hyberbola

    (p - ysin\alpha/cos\alpha )^2/a^2 - y^2/b^2 = 1

    (p^2 + y^2sin^2\alpha - 2pysin\alpha)/a^2 cos^2 \alpha  - y^2/b^2 = 1

    (p^2 + y^2sin^2\alpha - 2pysin\alpha)/a^2 cos^2 \alpha  =  1 + y^2/b^2

    (p^2 + y^2sin^2\alpha - 2pysin\alpha)b^2  =  (b^2 + y^2)(a^2 cos^2 \alpha)

    This doesn't seem to be leading anywhere. Is this what I am supposed to do?
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  10. #10
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    Re: Proof involving hyberbola

    Rearrange your last line to get a quadratic in y. You then want this quadratic to have a repeated root.
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  11. #11
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    Re: Proof involving hyberbola

    What you have substituted for x should be p/cosa-(sina/cosa)y
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  12. #12
    Don
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    Re: Proof involving hyberbola

    So I got the equation

    \left (p - y sin\alpha  \right )^2 = (a^2+b^2)(a^2cos^2\alpha )/b^2

    Is this correct?
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  13. #13
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    Re: Proof involving hyberbola

    I can't see what's happened there. Looking back your last line of post 9 was correct so look at my post 10 again
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    Re: Proof involving hyberbola

    If you want one last fresh start try this. I'll write A instead of alpha. Line can be written y=-(cosA/sinA)x+p/sinA
    This is y=mx+c with m=-cosA/sinA and c=p/sinA
    Substitute y=mx+c into the hyperbola. So x^2/a^2-(mx+c)^2/b^2=1
    Multiply throughout by a^2b^2 and remove the brackets and then rearrange as a quadratic in x
    Apply the condition for this to have repeated roots
    After some symplifying this gives b^2+c^2-(a^2m^2)=0
    Put back in what m and c stood for to get the required result
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  15. #15
    Don
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    Re: Proof involving hyberbola

    OK, here's my attempt:

    x^2/a^2 - (mx + c)^2/b^2 = 1

    Multiplying both the terms by a2b2,

    x^2b^2 - (m^2x^2 + c^2 + 2mxc)a^2 = 1

    Rearranging, I get

    x^2(b^2 - m^2a^2) + x(-2mca^2) + (-a^2c^2-1) = 0

    This is of the form ax2+bx+c=0

    What do I do now? I don't know how to make it have repeated roots.

    I hope I am on the right track.
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