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Finding tangent line that passes through a point not on curve

  1. Feb 23, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the number of tangent lines to the curve:

    [itex]y=\frac{3x}{x-2}[/itex]

    which pass through the point (-1,9). Find also the points of contact of these tangent lines with the curve.


    3. The attempt at a solution

    1. I found the equation of lines passing through (-1,9) -> [itex]y=(x+1)m+9[/itex]

    2. I thought there must be a contact point between the line through (-1,9) and the original equation so -> [itex]\frac{3x}{x-2}=(x+1)m+9[/itex]

    3. I put this into a quadratic form -> [itex]mx^{2}+(6-m)x-2m-18=0[/itex]

    4. I checked the discriminant of the equation in '3' -> [itex]9m^{2}+60m+36[/itex]

    5. I found the roots of '4' to be [itex]x=6[/itex] and [itex]x=-\frac{2}{3}[/itex] edit: [itex]m=6[/itex] and [itex]m=-\frac{2}{3}[/itex]

    I know that if the discriminant is 0 then there is one real solution. This means that there are two tangents of the original equation that also go through (-1,9).

    Now I don't know how to get the equation for these two tangent lines? How can I use the roots of the discriminant to get these equations?

    Any help appreciated!
     
    Last edited: Feb 23, 2013
  2. jcsd
  3. Feb 23, 2013 #2

    Dick

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    m is the slope. Why are you treating that as an unknown? That's the derivative dy/dx, yes? Put that it. The only real unknown here is x.
     
    Last edited: Feb 23, 2013
  4. Feb 23, 2013 #3
    Not sure I'm following what your saying. Do you mean that the roots of the discriminant i.e. X=6 and -2/3 are the slopes of the tangent lines I am looking for?
     
  5. Feb 23, 2013 #4

    Dick

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    I really don't know, you went off track a bit for me after step 2). Just put dy/dx in for m in step 2) and try to solve for x. That would be a correct thing to do, yes? dy/dx=m. The only real variable here is x.
     
    Last edited: Feb 23, 2013
  6. Feb 23, 2013 #5
    so you mean put in:

    [itex]m=\frac{-6}{(x-2)^{2}}[/itex]
     
  7. Feb 23, 2013 #6

    Dick

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    Indeed I do. Any reason why not?
     
  8. Feb 23, 2013 #7
    [itex]\frac{3x}{x-2}=\frac{(-6)(x+1)}{(x-2)(x-2)}+9[/itex]

    [itex]3x=\frac{(-6)(x+1)(x-2)}{(x-2)(x-2)}+9(x-2)[/itex]

    [itex]x=\frac{(-6)(x+1)(x-2)}{(3)(x-2)(x-2)}+\frac{9}{3}(x-2)[/itex]

    [itex]x=\frac{(-6)(x+1)}{(3)(x-2)}+3(x-2)[/itex]

    [itex]x=\frac{(-2)(x+1)}{(x-2)}+3(x-2)[/itex]

    [itex]x=\frac{-2x-2}{(x-2)}+3(x-2)[/itex]

    Cant seem to solve for X am I doing something wrong?
     
  9. Feb 23, 2013 #8

    Dick

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    Why don't you multiply by (x-2)^2 to begin with? You'll just get a quadratic equation to solve.
     
  10. Feb 23, 2013 #9
    Oh right like this... what is the next step?

    [itex]3(x-2)^{2}-x(x-2)-2x-2=0[/itex]
     
  11. Feb 23, 2013 #10
    [itex]2x^{2}-12x+10=0[/itex]

    [itex]2(x^{2}-6x+5)[/itex]

    [itex]2(x-1)(x-5)[/itex]

    so x=1,5???
     
  12. Feb 23, 2013 #11

    Dick

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    If you did that right, then multiply it out and solve the quadratic. If you didn't do it right then fix what's wrong and do it correctly. I'm too tired to check it right now. Good luck!
     
  13. Feb 23, 2013 #12

    Dick

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    Yes, x=1,5. Then you can use the x values to find m. So you have your two tangent lines.
     
  14. Feb 23, 2013 #13
    Great thanks Dick
     
  15. Feb 23, 2013 #14
    Oh wow the slopes of the tangents turn out to be:

    m=-2/3 and m=6

    Just as in the beginning!

    I.e. the roots of the discriminant
     
  16. Feb 23, 2013 #15

    Dick

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    Ok, that's good. You did understand the problem. I just took the more direct approach of eliminating m and then not looking at anything after that. Feel good?
     
  17. Feb 23, 2013 #16
    Yep lol. But I don't really understand why the root of the discriminant is the slope of the tangent line as a concept.
     
  18. Feb 24, 2013 #17

    Dick

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    It's because the tangent line intersects the curve exactly once. It's a trick to find tangent slopes without using calculus or derivatives. But I know you do know derivatives. It's a bit more straightforward if you use that.
     
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