Let A (4, 0, 0) & B (– 4, 0, 0)
Let the coordinates of point P be (x, y, z)
Calculating PA
P ≡ (x, y, z) and A ≡ (4, 0, 0)
By using the formula,
Distance PA =
\( \sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}\)
So here,
x1 = x, y1 = y, z1 = z
x2 = 4, y2 = 0, z2 = 0
Distance PA =
\( \sqrt{(4– x)^2 + (0 – y)^2 + (0 – z)^2}\)
Calculating PB
P ≡ (x, y, z) and B ≡ (– 4, 0, 0)
By using the formula,
Distance PB
= \( \sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}\)
So here,
x1 = x, y1 = y, z1 = z
x2 = – 4, y2 = 0, z2 = 0
Distance PB
= \( \sqrt{(4– x)^2 + (0 – y)^2 + (0 – z)^2}\)
Now it is given that:
PA + PB = 10
PA = 10 – PB
Square on both the sides, we get
PA2 = (10 – PB)2
PA2 = 100 + PB2 – 20 PB
(4 – x)2 + (0 – y)2 + (0 – z)2
100 + (– 4 – x)2 + (0 – y)2 + (0 – z)2 – 20 PB
(16 + x2 – 8x) + (y2) + (z2)
100 + (16 + x2 + 8x) + (y2) + (z2) – 20 PB
20 PB = 16x + 100
5 PB = (4x + 25)
Square on both the sides again, we get
25 PB2 = 16x2 + 200x + 625
25 [(– 4 – x)2 + (0 – y)2 + (0 – z)2] = 16x2 + 200x + 625
25 [x2 + y2 + z2 + 8x + 16] = 16x2 + 200x + 625
25x2 + 25y2 + 25z2 + 200x + 400 = 16x2 + 200x + 625
9x2 + 25y2 + 25z2 – 225 = 0
The required equation is 9x2 + 25y2 + 25z2 – 225 = 0
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