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 &gt;0)

Question

Find the locus of a point p whose distances from two fixed points A, A&#039; are in a constant ratio 1:M, that is |PA| : |PA&#039;| = 1:M. ( M &amp;gt;0)

Jazmin Cameron · Accepted Answer

Let us suppose that P,A and A&#039; has a coordinate (x,y),(x1,y2) and (x3,y3) respectively.First find the distance between P and A and distance between Pand A&#039; 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&#039;|= 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&#039;| Msqrt((x^2-2x*x1+x1^2) +(y^2-2y*y1+y1^2)) = 1sqrt((x^2-2x*x2+x2^2) +(y^2-2y*y2+y2^2)) MSquaring 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) MCross MultiplyingM((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

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)

Answered question

Answer & Explanation

New Questions in Elementary geometry