LynFran
  • LynFran
Find the equation of the locus of the intersection of the lines below
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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LynFran
  • LynFran
\[y=mx+\sqrt{m^2+2}\]and \[y=-\frac{ 1 }{ m }x+\sqrt{\frac{ 1 }{ m^2 +2}}\]
LynFran
  • LynFran
@loser @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
I'm assuming you first equated the two right hand sides?

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LynFran
  • LynFran
yes was thinking the same ...then what?
jim_thompson5910
  • jim_thompson5910
try to isolate x in the equation
LynFran
  • LynFran
\[mx+\sqrt{m^2+2}=-\frac{ 1 }{ m }x+\sqrt{\frac{ 1 }{ m^2 }+2}\]
LynFran
  • LynFran
\[mx+\frac{ 1 }{ m }x=\sqrt{\frac{ 1 }{ m^2 }+2}-\sqrt{m^2+2}\]
jim_thompson5910
  • jim_thompson5910
then you can factor out x
LynFran
  • LynFran
\[x(m+\frac{ 1 }{ m })=\sqrt{\frac{ 1 }{ m^2 }+2}-\sqrt{m^2+2}\]
jim_thompson5910
  • jim_thompson5910
then you can divide both sides by m+1/m
jim_thompson5910
  • jim_thompson5910
hmm the question is now: how to use this info to figure out the locus
LynFran
  • LynFran
\[x=\frac{ \sqrt{\frac{ 1 }{ m^2 }+2}-\sqrt{m^2+2} }{ m+\frac{ 1 }{ m } }\]
LynFran
  • LynFran
i dont know ...
LynFran
  • LynFran
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??
jim_thompson5910
  • jim_thompson5910
hmm that might be a possibility but I don't see where it would go in terms of being simplified
ganeshie8
  • ganeshie8
you want to eliminate \(m\)
LynFran
  • LynFran
but how do we eliminate that?? is m here the slope??
jim_thompson5910
  • jim_thompson5910
\[\Large y=mx+\sqrt{m^2+2}\] \[\Large y=m(\color{red}{x})+\sqrt{m^2+2}\] \[\Large y=m(\color{red}{\frac{ \sqrt{\frac{ 1 }{ m^2 }+2}-\sqrt{m^2+2} }{ m+\frac{ 1 }{ m } }})+\sqrt{m^2+2}\] idk where to go from there
LynFran
  • LynFran
im stuck there too
LynFran
  • LynFran
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...
jim_thompson5910
  • jim_thompson5910
idk what they mean by transpose. I'm used to the matrix definition shown here https://en.wikipedia.org/wiki/Transpose
jim_thompson5910
  • jim_thompson5910
maybe they mean take the reciprocal?
LynFran
  • LynFran
idk
jim_thompson5910
  • jim_thompson5910
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
LynFran
  • LynFran
ok i see ..go on
jim_thompson5910
  • jim_thompson5910
\[\Large y=-\frac{ 1 }{ m }x+\sqrt{\frac{ 1 }{ m^2 +2}}\] \[\Large my=-x+m\sqrt{\frac{ 1 }{ m^2 +2}}\] \[\Large x+my=m\sqrt{\frac{ 1 }{ m^2 +2}}\] \[\Large (x+my)^2=(m\sqrt{\frac{ 1 }{ m^2 +2}})^2\] \[\Large (x+my)^2=m^2(\frac{ 1 }{ m^2 +2})\]
LynFran
  • LynFran
\[(x+my)^2=m^2(\frac{ 1 }{ m^2 }+2)\]
jim_thompson5910
  • jim_thompson5910
\[\Large (x+my)^2=m^2(\frac{ 1 }{ m^2 +2})\] \[\Large \frac{(x+my)^2}{m^2}=\frac{ 1 }{ m^2 +2}\] \[\Large m^2 +2 = \frac{m^2}{(x+my)^2}\] ------------------------------------------------------- \[\Large (y-mx)^2=m^2+2\] \[\Large (y-mx)^2=\frac{m^2}{(x+my)^2}\] hmm not sure where to go from here
jim_thompson5910
  • jim_thompson5910
maybe take the square root of both sides to get \[\Large y-mx=\frac{m}{x+my}\]
jim_thompson5910
  • jim_thompson5910
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
LynFran
  • LynFran
\[(x+my)^2=1+2m^2\]
LynFran
  • LynFran
\[x^2+2xmy+(my)^2=1+2m^2\]idk
jim_thompson5910
  • jim_thompson5910
here's an applet to play around with if you're curious http://tube.geogebra.org/m/wlliSGAt
jim_thompson5910
  • jim_thompson5910
sliding around the m slider may be a bit slow, so be patient and let it load
jim_thompson5910
  • jim_thompson5910
as for how to eliminate m, I'm stumped. ganeshie8 may have an idea though
LynFran
  • LynFran
ok
triciaal
  • triciaal
according to the slopes the lines are perpendicular
triciaal
  • triciaal
do you have answer choices? using x = 0 then equating the y^2 I get y = 1 and y = -1
ganeshie8
  • ganeshie8
http://math.stackexchange.com/questions/1466618/find-the-equation-of-the-locus-of-the-intersection-of-the-lines-below

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