# Ex 12.2, 5 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 12.2, 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. Given A (4, 0, 0) B ( – 4, 0, 0) Let the Co-ordinate of point P be (x, y, z) Given that , Sum Distance of PA & PB is 10 i.e. PA + PB = 10 First, lets calculate PA & PB Calculating PA P (x, y, z) A (4, 0, 0) PA = x2−x12+y2−y12+z2 −z12 Here, x1 = x, y1 = y, z1 = z x2 = 4, y2 = 0, z2 = 0 PA = 4−𝑥2+0−𝑦2+0−𝑧2 = 4−𝑥2+−𝑦2+−𝑧2 = 4−𝑥2+𝑦2+𝑧2 Calculating PB P (x, y, z) B ( – 4, 0, 0) PB = x2−x12+y2−y12+z2 −z12 Here, x1 = x, y1 = y, z1 = z x2 = – 4, y2 = 0, z2 = 0 PB = −4−𝑥2+0−𝑦2+0−𝑧2 = −124+𝑥2+−𝑦2+−𝑧2 = 4+𝑥2+𝑦2+𝑧2 Now, given that PA + PB = 10 Putting value of PA & PB 4−𝑥2+y2+z2 + 4+𝑥2+y2+z2 = 10 4−𝑥2+y2+z2 = 10 – 4+𝑥2+y2+z2 Squaring both sides 4−𝑥2+y2+z2 = 10−4+𝑥2+y2+z22 (4 – x)2 + y2 + z2 = (10)2 + 4−𝑥2+y2+z22 – 2 (10) (4 – x)2 + y2 + z2 = 100 + (4 + x)2 + y2 + z2 – 204+𝑥2+y2+(z)2 (4 – x)2 –(4 + x)2 + y2 – y2 + z2 – z2 = – 204+𝑥2+y2+z2 (4 – x)2 – (4 + x)2 + 0 + 0 – 100 = – 204+𝑥2+y2+z2 ((4 – x) – (4 + x))((4 – x) + (4 + x)) – 100 = – 204+𝑥2+y2+z2 (4 – x – 4 + x)(4 – x + 4 + x) – 100 = – 204+𝑥2+y2+z2 ( – 2) (8) – 100 = – 204+𝑥2+y2+z2 – 16x – 100 – 20 4+𝑥2+y2+z2 – 4 (4x + 25) = – 4(5) 4+𝑥2+y2+z2 (x + 25) = – 54+𝑥2+y2+z2 Again Squaring both sides (x + 25)2 = ( – 5)2 4+𝑥2+y2+z22 (x + 25)2 = 25 ((4 + x)2 + y2 + z2) (4x + 25)2 = 25 ((4 + x)2 + y2 + z2) (4x)2 + (25)2 + 2(25)(x) = 25 ((4)2 + x2 + 2(4)(x) + y2 + z2 ) 16x2 + 625 + 200 x = 25(16 + x2 + 8x + y2 + z2 ) 16x2 + 625 + 200 x = 400 + 25x2 + 200x + 25y2 + 25z2 16x2 + 200 x – 25x2 – 200 x –25y2 – 25z2 = 400 – 625 16x2 – 25x2 – 25y2 – 25z2 – 200x + 200x = 225 – 9x2 –25x2 – 25y2 + 0 = 225 0 = 9x2 + 25x2 + 25y2 + 225 9x2 + 25x2 + 25y2 + 225 = 0 Which is the required equation of set of points P(x,y,z)

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