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Math Help - Hyperbola/Tangent proof

  1. #1
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    Hyperbola/Tangent proof

    I have been wrestling with this for about 45 mins now, I don't think it's that hard but I can't quite get it.

    Prove that the line with equation y=mx+c is a tangent to the hyperbola with equation \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 if, and only if a^2m^2=b^2+c^2

    Simple enough, right? But I ended up with two A4 sides of working (which I won't repeat here) and a complete mess. My approach was to prove that if it was a tangent then when I solved the two equations simultaneously the discriminant of the resulting quadratic would be equal to zero (since the tangent only touches once), I was hoping to rearrange this to get the required relationship between a,m,b and c. This didn't work as expected.

    I also have no idea how to prove the converse

    Any help would be great! Thanks!

    Last edited by Stonehambey; July 27th 2009 at 03:30 PM.
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  2. #2
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    Substituting:
    x^2 / a^2 - (mx + c)^2 / b^2 = 1

    Simplifying (if you can call it that!):
    x^2(a^2 m^2 - b^2) + 2 a^2 cmx + a^2(b^2 + c^2) = 0

    Substituting a^2 m^2 = b^2 + c^2 in 1st term, then b^2 + c^2 = a^2 m^2 in 3rd term:
    x^2 c^2 + 2 a^2 cmx + a^4 m^2 = 0

    x = [-2 a^2 cm +- SQRT(4 a^4 c^2 m^2 - 4 a^4 c^2 m^2)] / (2 c^2)
    There ya go: zero discriminant!

    x = -a^2 m / c

    I'm not saying I understand all that's going on... but I am proud to have
    arrived at a zero discriminant
    Last edited by Wilmer; July 27th 2009 at 09:02 PM. Reason: None
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  3. #3
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    Quote Originally Posted by Stonehambey View Post
    I have been wrestling with this for about 45 mins now, I don't think it's that hard but I can't quite get it.

    Prove that the line with equation y=mx+c is a tangent to the hyperbola with equation \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 if, and only if a^2m^2=b^2+c^2

    Simple enough, right? But I ended up with two A4 sides of working (which I won't repeat here) and a complete mess. My approach was to prove that if it was a tangent then when I solved the two equations simultaneously the discriminant of the resulting quadratic would be equal to zero (since the tangent only touches once), I was hoping to rearrange this to get the required relationship between a,m,b and c. This didn't work as expected.

    I also have no idea how to prove the converse

    Any help would be great! Thanks!

    Your approach looks OK to me. Post the highlights of your working.
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    Quote Originally Posted by mr fantastic View Post
    Your approach looks OK to me. Post the highlights of your working.
    Sure,

    y=mx+c and \frac{x^2}{a^2}-\frac{y^2}{b^2}=1

    \frac{x^2}{a^2}-\frac{(mx+c)^2}{b^2}=1

    \frac{x^2b^2-a^2(m^2x^2+2mcx+c^2)}{a^2b^2}=1
    .
    .
    .
    (b^2-a^2m^2)x^2-(2a^2mc)x-(c^2+a^2b^2)=0

    We require B^2-4AC=0, that is,

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

    I messed around with the above equation for a while before concluding that the mistake had already been made
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  5. #5
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    Quote Originally Posted by Stonehambey View Post
    Sure,

    y=mx+c and \frac{x^2}{a^2}-\frac{y^2}{b^2}=1

    \frac{x^2}{a^2}-\frac{(mx+c)^2}{b^2}=1

    \frac{x^2b^2-a^2(m^2x^2+2mcx+c^2)}{a^2b^2}=1
    .
    .
    .
    (b^2-a^2m^2)x^2-(2a^2mc)x-(c^2+a^2b^2)=0 Mr F says: This line is wrong. It should be (b^2-a^2m^2)x^2-(2a^2mc)x-({\color{red}a^2} c^2+a^2b^2)=0.

    [snip]
    Now get the discriminant and equate it to zero. Then expand and simplify.
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    Ah, thanks, that worked.

    So that shows "if tangent then relationship", does Wilmer's reply show "if relationship then tangent", since he subbed in the relationship to the equation and show that the discriminant was zero (so a tangent)?
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    Quote Originally Posted by mr fantastic View Post
    Now get the discriminant and equate it to zero. Then expand and simplify.
    MUCH easier to do the substituting first...as I did...
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  8. #8
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    Quote Originally Posted by Stonehambey View Post
    So that shows "if tangent then relationship", does Wilmer's reply show "if relationship then tangent", since he subbed in the relationship to the equation and show that the discriminant was zero (so a tangent)?
    Whoops...didn't see your post.
    Not sure; BUT I can't see any difference; like, who cares WHEN the
    substituting is done..
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  9. #9
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    Quote Originally Posted by Wilmer View Post
    Whoops...didn't see your post.
    Not sure; BUT I can't see any difference; like, who cares WHEN the
    substituting is done..
    Well, "p if and only if q" usually means you need to prove p => q AND q => p. Sometimes they're very similar, or one is trivial, but the two cases should still always be made clear
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