how to solve this locus equations? thanks

Question

P moves so that its distance from the origin equal its distance from the line x-y=3. Find the equation of its locus.

 

answers: x^2 +y^2 +2xy+6x-6y-9=0

Answer 1

Hello Valentine,

Find the distance of the point P from x-y=3, using dist. = (lx-y-3l)/√2

Dist. from Origin = √(x²+y²)

Equate the two distances and then simplify (squaring both sides helps).

The Answer is verified.

Of course the Locus will be a parabola with focus at the Origin and x-y=3 as the Directrix.

Pradip  

Answer 2

P(x,y), let O=(0,0), and A be a point on the line x-y=3. The condition is that PO=PA, the distance PO=sqrt(x^2+y^2) To find PA you must turn x-y=3 into standard form into x-y-3=0. PA=abs(x-y-3)/sqrt(2) square both sides of the equation, then x^2+y^2=(x-y-3)(x-y-3)/2, distribute and do the algebra, then you get x^2+y^2-2xy+6x-6y+9=0