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LynFran

  • one year ago

Find the equation of the locus of the intersection of the lines below

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  1. LynFran
    • one year ago
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    \[y=mx+\sqrt{m^2+2}\]and \[y=-\frac{ 1 }{ m }x+\sqrt{\frac{ 1 }{ m^2 +2}}\]

  2. LynFran
    • one year ago
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    @loser @jim_thompson5910

  3. jim_thompson5910
    • one year ago
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    I'm assuming you first equated the two right hand sides?

  4. LynFran
    • one year ago
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    yes was thinking the same ...then what?

  5. jim_thompson5910
    • one year ago
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    try to isolate x in the equation

  6. LynFran
    • one year ago
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    \[mx+\sqrt{m^2+2}=-\frac{ 1 }{ m }x+\sqrt{\frac{ 1 }{ m^2 }+2}\]

  7. LynFran
    • one year ago
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    \[mx+\frac{ 1 }{ m }x=\sqrt{\frac{ 1 }{ m^2 }+2}-\sqrt{m^2+2}\]

  8. jim_thompson5910
    • one year ago
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    then you can factor out x

  9. LynFran
    • one year ago
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    \[x(m+\frac{ 1 }{ m })=\sqrt{\frac{ 1 }{ m^2 }+2}-\sqrt{m^2+2}\]

  10. jim_thompson5910
    • one year ago
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    then you can divide both sides by m+1/m

  11. jim_thompson5910
    • one year ago
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    hmm the question is now: how to use this info to figure out the locus

  12. LynFran
    • one year ago
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    \[x=\frac{ \sqrt{\frac{ 1 }{ m^2 }+2}-\sqrt{m^2+2} }{ m+\frac{ 1 }{ m } }\]

  13. LynFran
    • one year ago
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    i dont know ...

  14. LynFran
    • one year ago
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    do we sub the value of x= blah blah lol (the one about ..dont feeling like typing all that back) into \[y=mx+\sqrt{m^2+2}\]to find y...idk??

  15. jim_thompson5910
    • one year ago
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    hmm that might be a possibility but I don't see where it would go in terms of being simplified

  16. ganeshie8
    • one year ago
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    you want to eliminate \(m\)

  17. LynFran
    • one year ago
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    but how do we eliminate that?? is m here the slope??

  18. jim_thompson5910
    • one year ago
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    \[\Large y=mx+\sqrt{m^2+2}\] \[\Large y=m(\color{red}{x})+\sqrt{m^2+2}\] \[\Large y=m(\color{red}{\frac{ \sqrt{\frac{ 1 }{ m^2 }+2}-\sqrt{m^2+2} }{ m+\frac{ 1 }{ m } }})+\sqrt{m^2+2}\] idk where to go from there

  19. LynFran
    • one year ago
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    im stuck there too

  20. LynFran
    • one year ago
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    there is this hint that says ..to eliminate m transpose the x term in each equation, clear the fraction, square both members and add .. but im still a bit confuse by what its saying...

  21. jim_thompson5910
    • one year ago
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    idk what they mean by transpose. I'm used to the matrix definition shown here https://en.wikipedia.org/wiki/Transpose

  22. jim_thompson5910
    • one year ago
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    maybe they mean take the reciprocal?

  23. LynFran
    • one year ago
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    idk

  24. jim_thompson5910
    • one year ago
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    Let me try something \[\Large y=mx+\sqrt{m^2+2}\] \[\Large y-mx=\sqrt{m^2+2}\] \[\Large (y-mx)^2=(\sqrt{m^2+2})^2\] \[\Large (y-mx)^2=m^2+2\] that gets rid of the pesky square root

  25. LynFran
    • one year ago
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    ok i see ..go on

  26. jim_thompson5910
    • one year ago
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    \[\Large y=-\frac{ 1 }{ m }x+\sqrt{\frac{ 1 }{ m^2 +2}}\] \[\Large my=-x+m\sqrt{\frac{ 1 }{ m^2 +2}}\] \[\Large x+my=m\sqrt{\frac{ 1 }{ m^2 +2}}\] \[\Large (x+my)^2=(m\sqrt{\frac{ 1 }{ m^2 +2}})^2\] \[\Large (x+my)^2=m^2(\frac{ 1 }{ m^2 +2})\]

  27. LynFran
    • one year ago
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    \[(x+my)^2=m^2(\frac{ 1 }{ m^2 }+2)\]

  28. jim_thompson5910
    • one year ago
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    \[\Large (x+my)^2=m^2(\frac{ 1 }{ m^2 +2})\] \[\Large \frac{(x+my)^2}{m^2}=\frac{ 1 }{ m^2 +2}\] \[\Large m^2 +2 = \frac{m^2}{(x+my)^2}\] ------------------------------------------------------- \[\Large (y-mx)^2=m^2+2\] \[\Large (y-mx)^2=\frac{m^2}{(x+my)^2}\] hmm not sure where to go from here

  29. jim_thompson5910
    • one year ago
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    maybe take the square root of both sides to get \[\Large y-mx=\frac{m}{x+my}\]

  30. jim_thompson5910
    • one year ago
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    I used geogebra to cheat and determined that the locus is actually the upper half of the ellipse \(\LARGE x^2 + 0.5y^2 = 1\). So solve that for y to get \(\LARGE y = \sqrt{2-2x^2}\) I just don't know how to eliminate m to get that

  31. LynFran
    • one year ago
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    \[(x+my)^2=1+2m^2\]

  32. LynFran
    • one year ago
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    \[x^2+2xmy+(my)^2=1+2m^2\]idk

  33. jim_thompson5910
    • one year ago
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    here's an applet to play around with if you're curious http://tube.geogebra.org/m/wlliSGAt

  34. jim_thompson5910
    • one year ago
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    sliding around the m slider may be a bit slow, so be patient and let it load

  35. jim_thompson5910
    • one year ago
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    as for how to eliminate m, I'm stumped. ganeshie8 may have an idea though

  36. LynFran
    • one year ago
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    ok

  37. triciaal
    • one year ago
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    according to the slopes the lines are perpendicular

  38. triciaal
    • one year ago
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    do you have answer choices? using x = 0 then equating the y^2 I get y = 1 and y = -1

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