NCERT Solutions
Class 11 Maths
Introduction to Three Dimensional Geometry
Ex.12.2 Q.5
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.
Let the coordinates of P be (x, y, z).
The coordinates of points A and B are (4, 0, 0) and (–4, 0, 0) respectively.
It is given that PA + PB = 10
=> √ {(x - 4)2 + y2 + z2} + √ {(x + 4)2 + y2 + z2} = 10
=> √ {(x - 4)2 + y2 + z2} = 10 - √ {(x + 4)2 + y2 + z2}
On squaring both sides, we obtain
=> (x - 4)2 + y2 + z2 = 102 + (x + 4)2 + y2 + z2 - 20√ {(x + 4)2 + y2 + z2}
=> x2 + 16 – 8x + y2 + z2 = 100 + x2 + 16 + 8x + y2 + z2 - 20√ {(x + 4)2 + y2 + z2}
=> 20√ {(x + 4)2 + y2 + z2} = 100 + 16x
=> 5√ {(x + 4)2 + y2 + z2} = 25 + 4x
On squaring both sides again, we obtain
=> [5√ {(x + 4)2 + y2 + z2}]2 = (25 + 4x)2
=> 25(x2 + 8x + 16 + y2 + z2) = 625 + 16x2 + 200x
=> 25x2 + 200x + 400 + 25y2 + 25z2 = 625 + 16x2 + 200x
=> 9x2 + 25y2 + 25z2 – 225 = 0
Thus, the required equation is 9x2 + 25y2 + 25z2 – 225 = 0