General Locus Proofs help (1 Viewer)

[karho]

hi
been overly fustrated on this... anyone help?

A and B are two fixed points and a point P(x,y) moves so that the distance PA is always double the distance PB. Choose suitable points for A and B and find that the equation of the locus proving in consquence that the locus is a circle whose centre is on the line AB. HInt: the final equation and all the algebra may be simpler if the orign (0,0) is on the locus. Remeber tat PA=2PB so think where it might be best to place A and B so that the origin is one point of the locus.

 

[Affinity]

hmm, better if you choosed wisely so that the origin becomes the center

let A be the point (4n,0) and B be the point (n,0) for some positive real n.

let P(x,y) satisfies:

|PA| = 2|PB|

|PA|^2 = 4|PB|^2

(x-4n)^2 + y^2 = 4[(x-n)^2 + y^2]

x^2 -8nx + 16n^2 + y^2 = 4x^2 - 8nx + 4n^2 + 4y^2

3x^2 + 3y^2 = 12n^2

x^2 + y^2 = (2n)^2

so .. it's a circle with radius 2/3 the distance between A and B, centered at the point which divides AB externally in the ratio of 1 to 4.

 

[Bepps]

--A be the point (4n,0) and B be the point (n,0) for some positive real n---

Why 4n?? If its twice the didtance wouldnt it be 2n? (Please explain in VERY simplistic terms, i'm terrible at the locus)

 

[Affinity]

because I know with that choice, the circle will be centered at origin and I can save myself doing algebra

I am just choosing the orgin so that it divides the line AB externally in ratio of 1 to 4