It three points A(-a,0),B(a,0) and P(x,y) are such that PA^2 + 2k^2, where k is constant show that x^2+y^2=k^2-a^2?
Answers
Given Point A=(a,0),B=(−a,0) for a moving point P
PA
2
−PB
2
=2k
2
for k is constant
also given to find out equation to the locus of point P
PA=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
P(x,y) A(a,0)
PA=
(a−x)
2
+(0−y)
2
by squaring on both sides for eqution (1) we get
(PA)
2
=[
(a−x)
2
+(−y)
2
]
2
(PA)
2
=a
2
+x
2
−2ax+y
2
(PA)
2
=x
2
+y
2
−2ax+a
2
x
2
+y
2
+a(a−2x)....(2)
PB=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
P(x,y);B(−a,0)
PB=
(−a−x)
2
+(0−y)
2
...(3)
by squaring on both side of (3) we get
(PB)
2
=[
(−a−x)
2
+(−y)
2
]
2
(PB)
2
=x
2
+2ax+a
2
+y
2
x
2
+y
2
+a(a+2x)....(4)
Let us substitute (PA)
2
and (PB)
2
in given equation
(PA)
2
−(PB)
2
=2k
2
x
2
+y
2
+a
2
−2ax−(x
2
+y
2
+a
2
+2ax)=2k
2
x
2
+y
2
+a
2
−2ax−x
2
−y
2
−a
2
−2ax=2k
2
−4ax=2k
2
k
2
=−2ax⇒ k
2
+2ax=0