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

1. LynFran

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

2. LynFran

@loser @jim_thompson5910

3. jim_thompson5910

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

4. LynFran

yes was thinking the same ...then what?

5. jim_thompson5910

try to isolate x in the equation

6. LynFran

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

7. LynFran

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

8. jim_thompson5910

then you can factor out x

9. LynFran

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

10. jim_thompson5910

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

11. jim_thompson5910

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

12. LynFran

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

13. LynFran

i dont know ...

14. 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??

15. jim_thompson5910

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

16. ganeshie8

you want to eliminate $$m$$

17. LynFran

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

18. 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

19. LynFran

im stuck there too

20. 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...

21. jim_thompson5910

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

22. jim_thompson5910

maybe they mean take the reciprocal?

23. LynFran

idk

24. 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

25. LynFran

ok i see ..go on

26. jim_thompson5910

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

27. LynFran

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

28. 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

29. jim_thompson5910

maybe take the square root of both sides to get $\Large y-mx=\frac{m}{x+my}$

30. 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

31. LynFran

$(x+my)^2=1+2m^2$

32. LynFran

$x^2+2xmy+(my)^2=1+2m^2$idk

33. jim_thompson5910

here's an applet to play around with if you're curious http://tube.geogebra.org/m/wlliSGAt

34. jim_thompson5910

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

35. jim_thompson5910

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

36. LynFran

ok

37. triciaal

according to the slopes the lines are perpendicular

38. triciaal

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

39. ganeshie8