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.
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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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