Find the equation of locus of a point P, if the distance of P from A (3,0) is twice the distance of P from B (-3,0).
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Answer:
Let the point P be (x,y)
The distance between P(x,y) and A(3,0) is
(3−x)
2
+(0−y)
2
The distance between P(x,y) and B(−3,0) is
(−3−x)
2
+(0−y)
2
Given that PA=2PB
⟹
(3−x)
2
+y
2
=2
(−3−x)
2
+y
2
squaring on both sides
⟹(3−x)
2
+y
2
=4((−3−x)
2
+y
2
)
⟹9+x
2
−6x+y
2
=4(9+x
2
+6x+y
2
)
⟹9+x
2
−6x+y
2
=36+4x
2
+24x+4y
2
⟹3x
2
+3y
2
+30x+27=0
Therefore, the locus of the point P is 3x
2
+3y
2
+30x+27=0
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