Question

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if A=(2,3) and B=(-2,1), find the equation of locus of point P such that AP^2=3BP^2

Answer 1

Let P=(x,y)
now AP=√{(x-2)^2+(y-3)^2
BP=√[{x-(-2)}^2+(y-1)^2
=√{(x+2)^2+(y-1)^2
since AP^2=3BP^2
we have (x-2)^2+(y-3)^2=3{(x+2)^2+(y-1)^2
Or, x^2-4x+4+y^2-6y+9=3(x^2+4x+4+y^2-2y+1)
Or, x^2-4x+y^2-6y+13=3(x^2+4x+y^2-2y+5)
Or, x^2-4x+y^2-6y+13=3x^2+12x+3y^2-6y+15
Or, x^2-3x^2+y^2-3y^2-4x-12x-6y+6y+13-15=0
Or, -2x^2-2y^2-16x-2=0
Or,-2(x^2+y^2+8x+1)=0
Or, x^2+y^2+8x+1=0
this is the required locus of P.