NCERT Class XI Mathematics - Introduction to Three Dimensional Geometry - Solutions

Question : 9

Total: 20

Find the equation of the set of points P, the sum of whose distances from A(4, 0, 0) and B(–4, 0, 0) is equal to 10.

Solution:

Let P(x, y, z) be any point.
Then PA = √(x−4)2+(y−0)2+(z−0)2 = √x2+16−8x+y2+z2
PB = √(x+4)2+(y−0)2+(z−0)2 = √x2+16+8x+y2+z2
It is given that PA + PB = 10
∴ √x2+16−8x+y2+z2 + √x2+16+8x+y2+z2 = 10
⇒ √x2+16−8x+y2+z2 = 10 - √x2+16+8x+y2+z2 ... (i)
Squaring (i) on both sides, we have
x2 + 16 - 8x + y2+z2 = 100 + x2 + 16 + 8x + y2+z2 - 20 √x2+16+8x+y2+z2
⇒ 20 √x2+16+8x+y2+z2 = 16x + 100
⇒ 5 = 4x + 25
√x2+16+8x+y2+z2
Squaring on both sides again, we have
25 (x2+16+8x+y2+z2) = 16x2 + 625 + 200x
⇒ 25x2 + 400 + 200x + 25y2+25z2–16x2 – 625 – 200x = 0
⇒ 9x2+25y2+25z2 – 225 = 0
Thus the required equation is 9x2+25y2+25z2 – 225 = 0

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