Question

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

Answer 1

\[y=mx+\sqrt{m^2+2}\]and \[y=-\frac{ 1 }{ m }x+\sqrt{\frac{ 1 }{ m^2 +2}}\]

Answer 2

@loser @jim_thompson5910

Answer 3

I'm assuming you first equated the two right hand sides?

Answer 4

yes was thinking the same ...then what?

Answer 5

try to isolate x in the equation

Answer 6

\[mx+\sqrt{m^2+2}=-\frac{ 1 }{ m }x+\sqrt{\frac{ 1 }{ m^2 }+2}\]

Answer 7

\[mx+\frac{ 1 }{ m }x=\sqrt{\frac{ 1 }{ m^2 }+2}-\sqrt{m^2+2}\]

Answer 8

then you can factor out x

Answer 9

\[x(m+\frac{ 1 }{ m })=\sqrt{\frac{ 1 }{ m^2 }+2}-\sqrt{m^2+2}\]

Answer 10

then you can divide both sides by m+1/m

Answer 11

hmm the question is now: how to use this info to figure out the locus

Answer 12

\[x=\frac{ \sqrt{\frac{ 1 }{ m^2 }+2}-\sqrt{m^2+2} }{ m+\frac{ 1 }{ m } }\]

Answer 13

i dont know ...

Answer 14

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??

Answer 15

hmm that might be a possibility but I don't see where it would go in terms of being simplified

Answer 16

you want to eliminate \(m\)

Answer 17

but how do we eliminate that?? is m here the slope??

Answer 18

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

Answer 19

im stuck there too

Answer 20

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...

Answer 21

idk what they mean by transpose. I'm used to the matrix definition shown here

https://en.wikipedia.org/wiki/Transpose

Answer 22

maybe they mean take the reciprocal?

Answer 23

idk

Answer 24

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

Answer 25

ok i see ..go on

Answer 26

\[\Large y=-\frac{ 1 }{ m }x+\sqrt{\frac{ 1 }{ m^2 +2}}\]
\[\Large my=-x+m\sqrt{\frac{ 1 }{ m^2 +2}}\]
\[\Large x+my=m\sqrt{\frac{ 1 }{ m^2 +2}}\]
\[\Large (x+my)^2=(m\sqrt{\frac{ 1 }{ m^2 +2}})^2\]
\[\Large (x+my)^2=m^2(\frac{ 1 }{ m^2 +2})\]

Answer 27

\[(x+my)^2=m^2(\frac{ 1 }{ m^2 }+2)\]

Answer 28

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

Answer 29

maybe take the square root of both sides to get

\[\Large y-mx=\frac{m}{x+my}\]

Answer 30

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

Answer 31

\[(x+my)^2=1+2m^2\]

Answer 32

\[x^2+2xmy+(my)^2=1+2m^2\]idk

Answer 33

here's an applet to play around with if you're curious

http://tube.geogebra.org/m/wlliSGAt

Answer 34

sliding around the m slider may be a bit slow, so be patient and let it load

Answer 35

as for how to eliminate m, I'm stumped. ganeshie8 may have an idea though

Answer 36

ok

Answer 37

according to the slopes the lines are perpendicular

Answer 38

do you have answer choices?
using x = 0 then equating the y^2 I get y = 1 and y = -1

Answer 39

http://math.stackexchange.com/questions/1466618/find-the-equation-of-the-locus-of-the-intersection-of-the-lines-below