# Misc 9

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Misc 9 Find the area of the smaller region bounded by the ellipse 𝑥2 𝑎2+ 𝑦2 𝑏2=1 & 𝑥𝑎 + 𝑦𝑏 = 1 Step 1: Drawing figure 𝒙𝟐 𝒂𝟐+ 𝒚𝟐 𝒃𝟐=𝟏 is an which is a equation ellipse with 𝑥−𝑎𝑥𝑖𝑠 as principle 𝑎𝑥𝑖𝑠 For 𝒙𝒂 + 𝒚𝒃 = 1 Points A(a, 0) and B(0, b) passes through both line and ellipse Required Area Required Area = Area OACB – Area OAB Area OACB Area OACB = 0𝑎𝑦 𝑑𝑥 𝑦 → Equation of ellipse 𝑥2 𝑎2+ 𝑦2 𝑏2=1 𝑦2 𝑏2=1− 𝑥2 𝑎2 𝑦2= 𝑏2 1− 𝑥2 𝑎2 𝑦= 𝑏2 1− 𝑥2 𝑎2 𝑦=𝑏 1− 𝑥2 𝑎2 Therefore, Area OACB = 0𝑎𝑏 1− 𝑥2 𝑎2𝑑𝑥 =b 0𝑎 𝑎2 − 𝑥2 𝑎2 𝑑𝑥 = 𝑏𝑎 0𝑎 𝑎2− 𝑥2 𝑑𝑥 = 𝑏𝑎 12𝑥 𝑎2− 𝑥2+ 𝑎22 sin−1 𝑥𝑎0𝑎 = 𝑏𝑎 12.𝑎 𝑎2− 𝑎2+ 𝑎22 sin−1 𝑎𝑎− 12 0 𝑎2− 02+ 𝑎22 sin−1 0𝑎 = 𝑏𝑎 0+ 𝑎22. 𝒔𝒊𝒏−𝟏𝟏−0−0 = 𝑏𝑎 0+ 𝑎22. 𝝅𝟐 = 𝑏𝑎 𝑎22 𝜋2 = 𝜋 𝑎𝑏 4 Area OAB Area OAB = 0𝑎𝑦 𝑑𝑥 𝑦 → Equation of line 𝑥𝑎+ 𝑦𝑏=1 𝑦𝑏=1− 𝑥𝑎 𝑦=𝑏 1− 𝑥𝑎 Therefore, Area OAB = 0𝑎𝑏 1− 𝑥𝑎𝑑𝑥 = 𝑏 𝑥− 𝑥22𝑎0𝑎 = 𝑏 𝑎− 𝑎22𝑎− 0− 022𝑎 = 𝑏 𝑎− 𝑎2−0 = 𝑎𝑏2 ∴ Area Required = Area OACB – Area OAB = 𝜋 𝑎𝑏 4− 𝑎𝑏2 = 𝑎𝑏2 𝜋2−1 = 𝑎𝑏2 𝜋−22 = 𝒂𝒃𝟒 𝝅−𝟐 square units

Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.