# Example 8

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 8 In Fig AOBA is the part of the ellipse 9𝑥2+𝑦2=36 in the first quadrant such that OA = 2 and OB = 6. Find the area between the arc AB and the chord AB Area Required = Area ABC Area ABC = Area OACB – Area ∆ OAB Area OACB Area OACB = 02𝑦𝑑𝑥 Here, 𝑦 → ellipse equation 9𝑥2+𝑦2=36 𝑦2=36−9𝑥2 𝑦=± 36−9 𝑥2 As OACB is in 1st quadrant, 𝑦= 36−9 𝑥2 Area OACB = 02𝑦𝑑𝑥 = 02 36−9 𝑥2 = 02 9( 369− 𝑥2) = 02 9(4− 𝑥2) = 02 32( 22− 𝑥2) = 3 02 22− 𝑥2𝑑𝑥 = 3 𝑥2 22− 𝑥2+ 222 sin−1 𝑥202 = 3 22 22− 22+2 sin−1 22− 02 22− 02−2 sin−10 = 3 1 0+2 sin−11−0−0 = 3 0+2 sin−11 = 3 × 2 × 𝜋2 = 3𝜋 Area OAB Area OAB = 02𝑦𝑑𝑥 Here, 𝑦 → equation of line AB Equation of line between A(2, 0) & B(0, 6) is 𝑦 − 0𝑥 − 2= 6 − 00 − 2 𝑦𝑥−2= 6−2 𝑦𝑥−2=−3 𝑦=−3 𝑥−2 Area OBC = 02𝑦 𝑑𝑥 = 02−3 𝑥−2 𝑑𝑥 = – 3 02 𝑥−2 𝑑𝑥 = – 3 𝑥22 −2𝑥 02 = – 3 222−2 ×2− 022−2 ×0 = – 3 −2 = 6 Area Required = Area OACB – Area ∆ OAB =3𝜋−6

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.