# Find the locus of the point, the sum of whose distances from the points A(4, 0, 0) and B(-4, 0, 0) is equal to 10.

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Find the locus of the point, the sum of whose distances from the points A(4, 0, 0) and B(-4, 0, 0) is equal to 10.

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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: ⇒ 16x+ 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