Given: Points are A(3, 4, 5) and B(-1, 3, -7)
To find: the locus of point P which moves in such a way that PA2 + PB2 = 2k2
Let the required point P(x, y, z)
The distance between any two points (a, b, c) and (m, n, o) is given by,
\(\sqrt{(a−m)^2+(b−n)^2+(c−0)^2}\)
Therefore,
The distance between P(x, y, z) and A(3, 4, 5) is PA,
= \(\sqrt{(x+3)^2+(y-4)^2+(z-5)^2}\)
Distance between P(x, y, z) and B(-1, 3, -7) is PB,
= \(\sqrt{(x-(-1))^2+(y-3)^2+(z-(-7))^2}\)
= \(\sqrt{(x+1)^2+(y-3)^2+(z+7)^2}\)
According to question:
PA2 + PB2 = 2k2
⇒ (x – 3)2+ (y – 4)2 + (z – 5)2 + (x + 1)2 + (y – 3)2 + (z + 7)2 = 2k2
⇒ x2+ 9 – 6x + y2 + 16 – 8y + z2 + 25 – 10z + x2+ 1 + 2x + y2+ 9 – 6y + z2 + 49 + 14z = 2k 2
⇒ 2x2+ 2y2 + 2z2 – 4x – 14y + 4z + 109 = 2k2
⇒ 2x2+ 2y2 + 2z2 – 4x – 14y + 4z + 109 – 2k2 = 0
Hence locus of point P is 2x2+ 2y2 + 2z2 – 4x – 14y + 4z + 109 – 2k2 = 0
