Help with locus qn. (1 Viewer)

Getteral09

Member
Joined
Apr 20, 2006
Messages
60
Gender
Undisclosed
HSC
N/A
Hey can anyone solve this maths problem? Point P moves so that PA^2 + PB^2 = 4 where A = (3,-1) and B = (-5, 4). Find the equation of the locus of P.
Thanks,Getteral
 

Getteral09

Member
Joined
Apr 20, 2006
Messages
60
Gender
Undisclosed
HSC
N/A
Yeh. there is a mistake. For some reason you can't get 25-6=47.

So how can you get the maths problem: (Point P moves so that PA^2 + PB^2 = 4 where A = (3,-1) and B = (-5, 4). Find the equation of the locus of P. ) to equal to the answer : 2x^2+4x+2y^2-6y+47?
 
Last edited:

xlr8-crillz

Member
Joined
Dec 21, 2005
Messages
118
Gender
Male
HSC
2008
maybe u dont make 2 equations.

maybe it goes:

x^2+y^2-6x+2y+6 + x^2+10x + y^2 - 8y + +37=4

solve solve solve

and the answer is:
0= 2x^2+4x+2y^2-6y+ 47
 

Getteral09

Member
Joined
Apr 20, 2006
Messages
60
Gender
Undisclosed
HSC
N/A
xlr8-crillz said:
maybe u dont make 2 equations.

maybe it goes:

x^2+y^2-6x+2y+6 + x^2+10x + y^2 - 8y + +37=4

solve solve solve

and the answer is:
0= 2x^2+4x+2y^2-6y+ 47
Nope, xlr8-crillz . That working out will eventually make the equation smaller i.e.
0= 2x^2+4x+2y^2-6y+ 39. And the other method happen to get the answer to
2x^2+4x+2y^2 -6y+43=0 which is wrong. Therefore i think there is a mistake in Maths In Focus's answer.
 

xlr8-crillz

Member
Joined
Dec 21, 2005
Messages
118
Gender
Male
HSC
2008
sorry instead of:
x^2+y^2-6x+2y+6 + x^2+10x + y^2 - 8y + 37=4

it should have been:
x^2+y^2-6x+2y+10 + x^2+10x + y^2 - 8y + 41=4

i just typed it out wrong

and its like that above cause u werent meant to minus the 4 and make it all = 0 in your two smaller equations of PA and PB. but just to be sure, u probably u should get a pro to confirm the answer.
 
Last edited:

followme

Member
Joined
Feb 22, 2006
Messages
79
Gender
Male
HSC
2006
PA<sup>2</sup>+PB<sup>2</sup>=2<sup>2</sup>
let the coordinate of P be (x,y)
use the distance formula:
PA=√(x-3)<sup>2</sup>+(y+1)<sup>2 </sup>& PB=√(x+5)<sup>2</sup>+(y-4)<sup>2</sup>
PA<sup>2</sup>=(x-3)<sup>2</sup>+(y+1)<sup>2 </sup>& PB<sup>2</sup>=(x+5)<sup>2</sup>+(y-4)<sup>2</sup>
ie (x-3)<sup>2</sup>+(y+1)<sup>2</sup>+(x+5)<sup>2</sup>+(y-4)<sup>2</sup>=4

solve...

The locus of P is 2X<sup>2</sup>+4X+2Y<sup>2</sup>-6Y+47=0
 
Last edited:

jyu

Member
Joined
Nov 14, 2005
Messages
623
Gender
Male
HSC
2006
Getteral09 said:
Hey can anyone solve this maths problem? Point P moves so that PA^2 + PB^2 = 4 where A = (3,-1) and B = (-5, 4). Find the equation of the locus of P.
Thanks,Getteral
No such point P can exist in the real x-y space
because AB = sqrt[(3+5)^2 + (-1-4)^2] = sqrt(89).
.: PA^2 + PB^2 > 4.

:) :) :wave:
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top