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LynFran
 one year ago
Find the equation of the locus of the intersection of the lines below
LynFran
 one year ago
Find the equation of the locus of the intersection of the lines below

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LynFran
 one year ago
Best ResponseYou've already chosen the best response.0\[y=mx+\sqrt{m^2+2}\]and \[y=\frac{ 1 }{ m }x+\sqrt{\frac{ 1 }{ m^2 +2}}\]

LynFran
 one year ago
Best ResponseYou've already chosen the best response.0@loser @jim_thompson5910

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0I'm assuming you first equated the two right hand sides?

LynFran
 one year ago
Best ResponseYou've already chosen the best response.0yes was thinking the same ...then what?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0try to isolate x in the equation

LynFran
 one year ago
Best ResponseYou've already chosen the best response.0\[mx+\sqrt{m^2+2}=\frac{ 1 }{ m }x+\sqrt{\frac{ 1 }{ m^2 }+2}\]

LynFran
 one year ago
Best ResponseYou've already chosen the best response.0\[mx+\frac{ 1 }{ m }x=\sqrt{\frac{ 1 }{ m^2 }+2}\sqrt{m^2+2}\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0then you can factor out x

LynFran
 one year ago
Best ResponseYou've already chosen the best response.0\[x(m+\frac{ 1 }{ m })=\sqrt{\frac{ 1 }{ m^2 }+2}\sqrt{m^2+2}\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0then you can divide both sides by m+1/m

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0hmm the question is now: how to use this info to figure out the locus

LynFran
 one year ago
Best ResponseYou've already chosen the best response.0\[x=\frac{ \sqrt{\frac{ 1 }{ m^2 }+2}\sqrt{m^2+2} }{ m+\frac{ 1 }{ m } }\]

LynFran
 one year ago
Best ResponseYou've already chosen the best response.0do 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
 one year ago
Best ResponseYou've already chosen the best response.0hmm that might be a possibility but I don't see where it would go in terms of being simplified

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0you want to eliminate \(m\)

LynFran
 one year ago
Best ResponseYou've already chosen the best response.0but how do we eliminate that?? is m here the slope??

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0\[\Large y=mx+\sqrt{m^2+2}\] \[\Large y=m(\color{red}{x})+\sqrt{m^2+2}\] \[\Large y=m(\color{red}{\frac{ \sqrt{\frac{ 1 }{ m^2 }+2}\sqrt{m^2+2} }{ m+\frac{ 1 }{ m } }})+\sqrt{m^2+2}\] idk where to go from there

LynFran
 one year ago
Best ResponseYou've already chosen the best response.0there 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
 one year ago
Best ResponseYou've already chosen the best response.0idk what they mean by transpose. I'm used to the matrix definition shown here https://en.wikipedia.org/wiki/Transpose

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0maybe they mean take the reciprocal?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0Let me try something \[\Large y=mx+\sqrt{m^2+2}\] \[\Large ymx=\sqrt{m^2+2}\] \[\Large (ymx)^2=(\sqrt{m^2+2})^2\] \[\Large (ymx)^2=m^2+2\] that gets rid of the pesky square root

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0\[\Large y=\frac{ 1 }{ m }x+\sqrt{\frac{ 1 }{ m^2 +2}}\] \[\Large my=x+m\sqrt{\frac{ 1 }{ m^2 +2}}\] \[\Large x+my=m\sqrt{\frac{ 1 }{ m^2 +2}}\] \[\Large (x+my)^2=(m\sqrt{\frac{ 1 }{ m^2 +2}})^2\] \[\Large (x+my)^2=m^2(\frac{ 1 }{ m^2 +2})\]

LynFran
 one year ago
Best ResponseYou've already chosen the best response.0\[(x+my)^2=m^2(\frac{ 1 }{ m^2 }+2)\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0\[\Large (x+my)^2=m^2(\frac{ 1 }{ m^2 +2})\] \[\Large \frac{(x+my)^2}{m^2}=\frac{ 1 }{ m^2 +2}\] \[\Large m^2 +2 = \frac{m^2}{(x+my)^2}\]  \[\Large (ymx)^2=m^2+2\] \[\Large (ymx)^2=\frac{m^2}{(x+my)^2}\] hmm not sure where to go from here

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0maybe take the square root of both sides to get \[\Large ymx=\frac{m}{x+my}\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0I 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 \(\LARGE y = \sqrt{22x^2}\) I just don't know how to eliminate m to get that

LynFran
 one year ago
Best ResponseYou've already chosen the best response.0\[x^2+2xmy+(my)^2=1+2m^2\]idk

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0here's an applet to play around with if you're curious http://tube.geogebra.org/m/wlliSGAt

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0sliding around the m slider may be a bit slow, so be patient and let it load

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0as for how to eliminate m, I'm stumped. ganeshie8 may have an idea though

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0according to the slopes the lines are perpendicular

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0do you have answer choices? using x = 0 then equating the y^2 I get y = 1 and y = 1
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