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Misc 8 - Find area of smaller region bounded by ellipse - Miscellaneous

Misc 8 - Chapter 8 Class 12 Application of Integrals - Part 2
Misc 8 - Chapter 8 Class 12 Application of Integrals - Part 3
Misc 8 - Chapter 8 Class 12 Application of Integrals - Part 4
Misc 8 - Chapter 8 Class 12 Application of Integrals - Part 5
Misc 8 - Chapter 8 Class 12 Application of Integrals - Part 6
Misc 8 - Chapter 8 Class 12 Application of Integrals - Part 7

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Misc 8 Find the area of the smaller region bounded by the ellipse 2 9 + 2 4 =1 & 3 + 2 = 1 Step 1: Drawing figure 2 9 + 2 4 =1 3 2 + 2 2 2 =1 Is an equation of an ellipse in the form 2 2 + 2 2 =1 with > which is a equation ellipse with as principle For + = Points A(2, 0) and B(0, 3) passes through both line and ellipse Required Area Required Area = Area OACB Area OAB Area OACB Area OACB = 0 3 Equation of ellipse 2 9 + 2 4 =1 2 4 =1 2 9 =4 1 2 9 = 4 1 2 9 =2 1 2 9 Therefore, Area OACB =2 0 3 1 2 9 =2 0 3 9 2 9 = 2 3 0 3 9 2 = 2 3 0 3 3 2 2 = 2 3 1 2 9 2 + 9 2 sin 1 3 0 3 = 2 3 1 2 .3 9 3 2 + 9 2 sin 1 3 3 1 2 0 9 0 2 + 9 2 sin 1 0 = 2 3 . 3 2 0 + 9 2 sin 1 1 0 0 = 2 3 0+ 9 2 . 2 = 3 2 Area OAB Area OAB = 0 3 Equation of line 3 + 2 =1 2 =1 3 =2 1 2 Therefore, Area OAB = 0 3 2 1 3 =2 0 3 1 3 =2 2 3 2 0 3 =2 2 6 0 3 =2 3 3 2 6 0 0 2 6 =2 3 3 2 =2. 3 2 =3 Thus, Required Area = Area OACB Area OAB = 3 2 3 = 3 2 1 = 3 2 2 square units

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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.