The coordinates of the points A and B are (a,0) and (−a,0), respectively. If a point P moves so that PA2−PB2=2k2, when k is constant, then find the equation to the locus of the point P.
2ax+k2=0
2ax−k2=0
ax+2k2=0
ax−2k2=0
A
2ax+k2=0
B
ax−2k2=0
C
2ax−k2=0
D
ax+2k2=0
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Solution
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Given Point A=(a,0),B=(−a,0) for a moving point P
PA2−PB2=2k2
for k is constant
also given to find out equation to the locus of point P
PA=√(x2−x1)2+(y2−y1)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=a2+x2−2ax+y2
(PA)2=x2+y2−2ax+a2
x2+y2+a(a−2x)....(2)
PB=√(x2−x1)2+(y2−y1)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=x2+2ax+a2+y2
x2+y2+a(a+2x)....(4)
Let us substitute (PA)2 and (PB)2 in given equation
(PA)2−(PB)2=2k2
x2+y2+a2−2ax−(x2+y2+a2+2ax)=2k2
x2+y2+a2−2ax−x2−y2−a2−2ax=2k2
−4ax=2k2
k2=−2ax⇒k2+2ax=0
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