# Example 5 - Chapter 8 Class 12 Application of Integrals

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 5 Find the area bounded by the ellipse 𝑥2 𝑎2+ 𝑦2 𝑏2=1 and the ordinates 𝑥=0 and 𝑥=𝑎𝑒, where, 𝑏2=𝑎2 (1 – 𝑒2) and e < 1 Required Area = Area of shaded region = Area BORQSP = 2 × Area OBPS = 2 × 0𝑎𝑒𝑦.𝑑𝑥 We know that , 𝑥2 𝑎2+ 𝑦2 𝑏2=1 𝑦2 𝑏2=1− 𝑥2 𝑎2 𝑦2 𝑏2= 𝑎2 − 𝑥2 𝑎2 𝑦2= 𝑏2 𝑎2 𝑎2− 𝑥2 ∴ 𝑦=± 𝑏2 𝑎2 𝑎2− 𝑥2 As OBPS is in 1st quadrant 𝑦= 𝑏𝑎 𝑎2− 𝑥2 Required Area = 2 × 0𝑎𝑒𝑦.𝑑𝑥 = 2 0𝑎𝑒 𝑏𝑎 𝑎2− 𝑥2 𝑑𝑥 = 2𝑏𝑎 0𝑎𝑒 𝑎2− 𝑥2 𝑑𝑥 = 2𝑏𝑎 12𝑥 𝑎2− 𝑥2+ 𝑎2𝑎 sin−1 𝑥𝑎0𝑎𝑒 = 2𝑏𝑎 𝑎𝑒2 𝑎2− 𝑎𝑒2+ 𝑎22 sin−1 𝑎𝑒𝑎− 02 𝑎2−0− 𝑎22 sin−1 0𝑎 = 2𝑏𝑎 𝑎𝑒2 𝑎2− 𝑎2 𝑒2+ 𝑎22 sin−1 𝑒−0− 𝑎22 sin−1 0 = 2𝑏𝑎 𝑎𝑒2.𝑎 1− 𝑒2+ 𝑎22 sin−1𝑒−0 = 2𝑏𝑎 𝑎2𝑒2 1− 𝑒2+ 𝑎22 sin−1𝑒 = 2𝑏𝑎 𝑎22 𝑒 1− 𝑒2+ sin−1𝑒 =𝑎𝑏 𝑒 1− 𝑒2+ sin−1𝑒 ∴ Required Area =𝒂𝒃 𝒆 𝟏− 𝒆𝟐+ 𝒔𝒊𝒏−𝟏𝒆 square units

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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.