Question

2. Problem: 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).​

Answer 1

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

Step-by-step explanation:

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Answer 2

Answer:

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