Find the locus of a point p whose distances from two fixed points A, A' are in a constant ratio 1:M, that is |PA| : |PA'| = 1:M. ( M >0)

Trevor Rush 2022-08-08 Answered
Find the locus of a point p whose distances from two fixed points A, A' are in a constant ratio 1:M, that is |PA| : |PA'| = 1:M. ( M >0)
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Answers (1)

Jazmin Cameron
Answered 2022-08-09 Author has 16 answers
Let us suppose that P,A and A' has a coordinate (x,y),(x1,y2) and (x3,y3) respectively.
First find the distance between P and A and distance between Pand A' through distance formula.
DISTANCE B/W Pand A:
|PA|= sqrt((x1-x)^2+(y1-y)^2)
=sqrt((x^2-2x*x1+x1^2) +(y^2-2y*y1+y1^2))
|PA'|= sqrt((x2-x)^2+(y2-y)^2)
=sqrt((x^2-2x*x2+x2^2) +(y^2-2y*y2+y2^2))
Now Applying condition:
|PA| = 1
|PA'| M
sqrt((x^2-2x*x1+x1^2) +(y^2-2y*y1+y1^2)) = 1
sqrt((x^2-2x*x2+x2^2) +(y^2-2y*y2+y2^2)) M
Squaring on both sides
(x^2-2x*x1+x1^2) +(y^2-2y*y1+y1^2) = 1
(x^2-2x*x2+x2^2) +(y^2-2y*y2+y2^2) M
Cross Multiplying
M((x^2-2x*x1+x1^2) +(y^2-2y*y1+y1^2)) = (x^2-2x*x2+x2^2) +(y^2-2y*y2+y2^2)
M*x^2-2M*x*x1+M*x1^2+M*y^2-2*M*y*y1+M*y1^2 = (x^2-2x*x2+x2^2) +(y^2-2y*y2+y2^2)
M*x^2-x^2+M*y^2-y^2-2M*x*x1+2x*x2-2*M*y*y1+2y*y2+M*x1^2+M*y1^2-x2^2-y2^2=0
(M-1)x^2+(M-1)y^2-2(M*x1-2*x2)x-2(M*y1-2*y2)y+(M*x1^2+M*y1^2-x2^2-y2^2)=0
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