Last updated at March 9, 2017 by Teachoo

Transcript

Example 13 Find the equation of the ellipse, with major axis along the x-axis and passing through the points (4, 3) and (– 1,4). Given that Major axis is along x-axis So required equation of ellipse is 𝒙𝟐𝒂𝟐 + 𝒚𝟐𝒃𝟐 = 1 Given that point (4, 3) & (−1, 4) lie of the ellipse So, point (4, 3) & (−1, 4) will satisfy equation of ellipse Putting x = 4 & y = 3 in (1) 𝑥2𝑎2 + 𝑦2𝑏2 = 1 (4)2𝑎2 + (3)2𝑏2 = 1 𝟏𝟔𝒂𝟐 + 𝟗𝒃𝟐 = 1 Putting x = −1, y = 4 is in (1) 𝑥2𝑎2 + 𝑦2𝑏2 = 1 (−1)2𝑎2 + (4)2𝑏2 = 1 𝟏𝒂𝟐 + 𝟏𝟔𝒃𝟐 = 1 Now, our equations are 16𝑎2 + 9𝑏2 = 1 …(2) 1𝑎2 + 16𝑏2 = 1 …(3) From (3) 1𝑎2+16𝑏2 = 1 1𝑎2 = 1−16𝑏2 Putting value of 1𝑎2 in (2) 16𝑎2 + 9𝑏2 = 1 161𝑎2 + 9𝑏2 = 1 161−16𝑏2 + 9𝑏2 = 1 16 − 256𝑏2 + 9𝑏2 = 1 −256 + 9𝑏2 = 1 −16 −247𝑏2 = −15 b2 = −247−15 b2 = 𝟐𝟒𝟕𝟏𝟓 Putting value of b2 = 24715 in (3) 1𝑎2 = 1−16𝑏2 1𝑎2 = 1−1624715 1𝑎2 = 1−16 × 15247 1𝑎2 = 247 − 240247 1𝑎2 = 7247 𝐚𝟐 = 𝟐𝟒𝟕𝟕 Thus, a2 = 2477 & b2 = 24715 Hence required of ellipse is 𝑥2𝑎2 + 𝑦2𝑏2 = 1 Putting values of a2 & b2 𝑥22477 + 𝑦224715 = 1 7𝑥2247 + 15𝑦2247 = 1 7x2 + 15y2 = 247

Class 11

Important Question for exams Class 11

- Chapter 1 Class 11 Sets
- Chapter 2 Class 11 Relations and Functions
- Chapter 3 Class 11 Trigonometric Functions
- Chapter 4 Class 11 Mathematical Induction
- Chapter 5 Class 11 Complex Numbers
- Chapter 6 Class 11 Linear Inequalities
- Chapter 7 Class 11 Permutations and Combinations
- Chapter 8 Class 11 Binomial Theorem
- Chapter 9 Class 11 Sequences and Series
- Chapter 10 Class 11 Straight Lines
- Chapter 11 Class 11 Conic Sections
- Chapter 12 Class 11 Introduction to Three Dimensional Geometry
- Chapter 13 Class 11 Limits and Derivatives
- Chapter 14 Class 11 Mathematical Reasoning
- Chapter 15 Class 11 Statistics
- Chapter 16 Class 11 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.