Question

A(3,0) and B(-3,0) are two fixed points. Find the locus of a moving point P such that PA is perpendicular to PB.​

Answer 1

Answer:

Given, A=(−3,0),B=(3,0) and PA=nPB

Let P be (h,k)

PA=(h+3,k)

PB=(h−3,k)

So, PA=nPB

⇒(h+3)

2

+k

2

=n

2

[(h−3)

2

+k

2

]

∴ locus of P(h,k) is

⇒ x

2

+6x+9+y

2

=n

2

[x

2

−6x+9+y

2

]

⇒ x

2

(1−n

2

)+y

2

(1−n

2

)+6x(1+n

2

)+9(1−n

2

)=0

⇒ x

2

+y

2

+6

1−n

2

(1+n

2

)

x+9=0(∵n



=1)

∴ locus is a circle

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