If A (3,0), B(-3,0) then the locus of the point P is such that PA^2+PB^2=18 is About the author Mary
Step-by-step explanation: let P(X, Y) If A (3,0), B(-3,0) then the locus of the point P is such that PA^2+PB^2=18 is [tex] ({x – 3})^{2} + {y}^{2} + {(x + 3)}^{2} + {y}^{2} = 18[/tex] [tex] {x}^{2} – 6x + 9 + {y}^{2} + {x}^{2} + 6x + 9 + {y}^{2} = 18[/tex] [tex]2 {x}^{2} + 2 {y}^{2} = 0[/tex] [tex] {x}^{2} + {y}^{2} = 0[/tex] Reply
Step-by-step explanation:
let P(X, Y)
If A (3,0), B(-3,0) then the locus of the point P is such that PA^2+PB^2=18 is
[tex] ({x – 3})^{2} + {y}^{2} + {(x + 3)}^{2} + {y}^{2} = 18[/tex]
[tex] {x}^{2} – 6x + 9 + {y}^{2} + {x}^{2} + 6x + 9 + {y}^{2} = 18[/tex]
[tex]2 {x}^{2} + 2 {y}^{2} = 0[/tex]
[tex] {x}^{2} + {y}^{2} = 0[/tex]