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Ex 12.2, 5 - Find equation of set of points P, the sum - Ex 12.2

Ex 12.2,  5 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 2
Ex 12.2,  5 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 3 Ex 12.2,  5 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 4 Ex 12.2,  5 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 5 Ex 12.2,  5 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 6


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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 coordinates of point P be (x, y, z) Given that , Sum Distance of PA & PB is 10 PA + PB = 10 First, lets calculate PA & PB Calculating PA P (x, y, z), A (4, 0, 0) PA = √((x2−x1)2+(y2−y1)2+(z2 −z1)2) Here, x1 = x, y1 = y, z1 = z x2 = 4, y2 = 0, z2 = 0 PA = √((4−𝑥)2+(0−𝑦)2+(0−𝑧)2) = √((4−𝑥)2+(−𝑦)2+(−𝑧)2) = √((4−𝑥)2+𝑦2+𝑧2) Calculating PB P (x, y, z), B ( – 4, 0, 0) PB = √((x2−x1)2+(y2−y1)2+(z2 −z1)2) Here, x1 = x, y1 = y, z1 = z x2 = – 4, y2 = 0, z2 = 0 PB = √((−4−𝑥)2+(0−𝑦)2+(0−𝑧)2) = √((−1)2(4+𝑥)2+(−𝑦)2+(−𝑧)2) = √((4+𝑥)2+𝑦2+𝑧2) Now, given that PA + PB = 10 Putting value of PA & PB √((4−𝑥)2+y2+z2) + √((4+𝑥)2+y2+z2) = 10 √((4−𝑥)2+y2+z2) = 10 – √((4+𝑥)2+y2+z2) Squaring both sides (4−𝑥)2+y2+z2 = (10−√((4+𝑥)2+y2+z2))^2 (4 – x)2 + y2 + z2 = (10)2 + (√((4+𝑥)2+y2+z2))2 – 2 (10) √((4+𝑥)2+y2+(z)2) (4 – x)2 + y2 + z2 = 100 + (4 + x)2 + y2 + z2 – 20√((4+𝑥)2+y2+(z)2) (4 – x)2 – (4 + x)2 + y2 – y2 + z2 – z2 = –20√((4+𝑥)2+y2+z2) (4 – x)2 – (4 + x)2 + 0 + 0 – 100 = –20√((4+𝑥)2+y2+z2) ((4 – x) – (4 + x))((4 – x) + (4 + x)) – 100 = –20√((4+𝑥)2+y2+z2) (4 – x – 4 − x)(4 – x + 4 + x) – 100 = –20√((4+𝑥)2+y2+z2) (–2x) (8) – 100 = –20√((4+𝑥)2+y2+z2) –16x – 100 = –20 √((4+𝑥)2+y2+z2) –4(4x + 25) = –4(5) √((4+𝑥)2+y2+z2) (4x + 25) = –5√((4+𝑥)2+y2+z2) Again Squaring both sides (4x + 25)2 = 25 ((4 + x)2 + y2 + z2) (4x)2 + (25)2 + 2(25)(4x) = 25 ((4)2 + x2 + 2(4)(x) + y2 + z2 ) 16x2 + 625 + 200x = 25(16 + x2 + 8x + y2 + z2) 16x2 + 625 + 200x = 400 + 25x2 + 200x + 25y2 + 25z2 16x2 + 200x – 25x2 – 200 x –25y2 – 25z2 = 400 – 625 16x2 – 25x2 – 25y2 – 25z2 – 200x + 200x = –225 – 9x2 –25y2 – 25z2 + 0 = –225 0 = 9x2 + 25y2 + 25z2 – 225 9x2 + 25y2 + 25z2 – 225 = 0 Which is the required equation of set of points P(x, y, z)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.