Question

Find the locus of a point such that the difference of the square of its distance from the point A(3, 4,5) and B(-1, 3,-7) is equal to 2k*2
​

Answer 1

Answer:

PA

2

−PB

2

=2k

2

⇒[(x−3)

2

+(y−4)

2

+(z−5)

2

]−[(x+1)

2

+(y−3)

2

+(z+7)

2

]=2k

2

⇒(x−3+x+1)(x−3−x−1)+(y−4+y−3)(y−4−y+3)+(z−5+z+7)(z−5−z−7)=2k

2

⇒(2x−2)(−4)+(2y−7)(−1)+(2z+1)(−12)=2k

2

⇒−8x+8−2y+7−24z−24=2k

2

⇒−8x−2y−24z−9=2k

2

⇒8x+2y+24z+9+2k

2

=0

Hope it helps you