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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Ex 8.1, 1 Find the area of the region bounded by the ellipse π‘₯^2/16+𝑦^2/9=1Equation Of Given Ellipse is π‘₯^2/16+𝑦^2/9=1 𝒙^𝟐/(πŸ’)^𝟐 +π’š^𝟐/(πŸ‘)^𝟐 =𝟏 Area of ellipse = Area of ABCD = 2 Γ— [Area Of ABC] = 2 Γ— ∫_(βˆ’πŸ’)^πŸ’β–’γ€–π’š.γ€— 𝒅𝒙 Finding y We know that π‘₯^2/16+𝑦^2/9=1 𝑦^2/9=1βˆ’π‘₯^2/16 𝑦^2/9=(16βˆ’π‘₯^2)/16 π’š^𝟐=πŸ—/πŸπŸ” (πŸπŸ”βˆ’π’™^𝟐 ) Taking square root on both sides y = Β± √(9/16 (16βˆ’π‘₯^2 ) ) y = Β± 3/4 √(16βˆ’π‘₯^2 ) Since, ABC is above x-axis y will be positive ∴ π’š=πŸ‘/πŸ’ √(πŸπŸ”βˆ’π’™^𝟐 ) Now, Area of ellipse = 2 Γ— ∫_(βˆ’4)^4▒〖𝑦.γ€— 𝑑π‘₯ = 2 Γ— ∫_(βˆ’πŸ’)^πŸ’β–’γ€– πŸ‘/πŸ’ √(πŸπŸ”βˆ’π’™^𝟐 )γ€— 𝒅𝒙 = 2 Γ— 3/4 ∫_(βˆ’4)^4β–’βˆš(16βˆ’π‘₯^2 ) 𝑑π‘₯ = πŸ‘/𝟐 ∫_(βˆ’πŸ’)^πŸ’β–’βˆš((πŸ’)^πŸβˆ’π’™^𝟐 ) 𝒅𝒙 It is of form √(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯=1/2 π‘₯√(π‘Ž^2βˆ’π‘₯^2 )+π‘Ž^2/2 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 π‘₯/π‘Ž+𝑐〗 Replacing a by 4 we get = 3/2 [π‘₯/2 √((4)^2βˆ’π‘₯^2 )+(4)^2/2 sin^(βˆ’1)⁑〖 π‘₯/4γ€— ]_(βˆ’4)^4 = 3/2 [4/2 √((4)^2βˆ’(4)^2 )βˆ’((βˆ’4))/2 √((4)^2βˆ’(βˆ’4)^2 )+16/2 γ€– sinγ€—^(βˆ’1)⁑〖(4/4)βˆ’16/2γ€— sin^(βˆ’1) ((βˆ’4)/4)] = 3/2 [2(0)+2(0)+8 γ€–sin^(βˆ’1) (1)γ€—β‘γ€–βˆ’ 8 sin^(βˆ’1)⁑(βˆ’1) γ€— ] = 3/2 [0+8 sin^(βˆ’1)⁑〖(1)βˆ’8 γ€–π’”π’Šπ’γ€—^(βˆ’πŸ)⁑(βˆ’πŸ) γ€— ] = 3/2 [8 sin^(βˆ’1)⁑〖(1)βˆ’8(βˆ’γ€–π’”π’Šπ’γ€—^(βˆ’πŸ)⁑(𝟏))γ€— ] = 3/2 [8 sin^(βˆ’1)⁑〖(1)+8 sin^(βˆ’1)⁑(1) γ€— ] = 3/2 Γ— 16 γ€–π’”π’Šπ’γ€—^(βˆ’πŸ)⁑(𝟏) = 3/2 Γ— 16 Γ— 𝝅/𝟐 = 3 Γ— 8 Γ— πœ‹/2 = 12Ο€ ∴ Area of Ellipse = 12Ο€ Square units

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.