Given: Points are A(4, 0, 0) and B(-4, 0, 0)
To find: the locus of point P, the sum of whose distances from the given points is equal to 10, i.e. PA + PB = 10
Let the required point P(x, y, z)
Formula used:
The distance between any two points (a, b, c) and (m, n, o) is given by,
\(\sqrt{(a−m)^2+(b−n)^2+(c−o)^2}\)
The distance between P(x, y, z) and A(4, 0, 0) is PA,
=\(\sqrt{(x-4)^2+(y−0)^2+(z-0)^2}\)
=\(\sqrt{(x-4)^2+y^2+z^2}\)
Distance between P(x, y, z) and B(-4, 0, 0) is PB
=\(\sqrt{(x-(-4))^2+(y−0)^2+(z-0)^2}\)
= \(\sqrt{(x-3)^2+(y−2)^2+(z+1)^2}\)
=\(\sqrt{(x+4)^2+y^2+z^2}\)
According to question: PA + PB = 10
⇒ \(\sqrt{(x-4)^2+y^2+z^2}\) + \(\sqrt{(x+4)^2+y^2+z^2}\) = 10
⇒ \(\sqrt{(x-4)^2+y^2+z^2}\) =10- \(\sqrt{(x+4)^2+y^2+z^2}\)
Squaring both sides:
⇒ 16x2 + 625 – 100x = 25x2+ 400 + 200x + 25y2 + 25z2
⇒ 16x2+ 625 – 100x – 25x2– 400 – 200x – 25y2 – 25z2 = 0
⇒ -9x2 – 25y2 – 25z2 – 300x + 225 = 0
⇒ 9x2 + 25y2 + 25z2 + 300x – 225 = 0
Hence locus of point P is 9x2 + 25y2 + 25z2 + 300x – 225 = 0
