Example 8 - AOBA is part of ellipse 9x2 + y2 = 36, OA = 2 - Examples

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  1. Chapter 8 Class 12 Application of Integrals
  2. Serial order wise
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Example 8 In Fig AOBA is the part of the ellipse 9 2+ 2=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 = 0 2 Here, ellipse equation 9 2+ 2=36 2=36 9 2 = 36 9 2 As OACB is in 1st quadrant, = 36 9 2 Area OACB = 0 2 = 0 2 36 9 2 = 0 2 9( 36 9 2 ) = 0 2 9(4 2 ) = 0 2 3 2 ( 2 2 2 ) = 3 0 2 2 2 2 = 3 2 2 2 2 + 2 2 2 sin 1 2 0 2 = 3 2 2 2 2 2 2 +2 sin 1 2 2 0 2 2 2 0 2 2 sin 1 0 = 3 1 0 +2 sin 1 1 0 0 = 3 0+2 sin 1 1 = 3 2 2 = 3 Area OAB Area OAB = 0 2 Here, equation of line AB Equation of line between A(2, 0) & B(0, 6) is 0 2 = 6 0 0 2 2 = 6 2 2 = 3 = 3 2 Area OBC = 0 2 = 0 2 3 2 = 3 0 2 2 = 3 2 2 2 0 2 = 3 2 2 2 2 2 0 2 2 2 0 = 3 2 = 6 Area Required = Area OACB Area OAB =3 6

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